BOWSER'S  MATHEMATICS. 


ACADEMIC  ALGEBRA.    With  numerous  Examples. 
COLLEGE  ALGEBRA.    With  numerous  Examples. 
PLANE  AND  SOLID  GEOMETRY.     With  numerous  Exer- 
cises. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC  GEOMETRY, 
embracing  Plane  Geometry,  and  au  Introduction  to 
Geometry  of  Three  Dimensions. 

AN  ELEMENTARY  TREATISE  ON  THE  DIFFERENTIAL 
AND  INTEGRAL  CALCULUS.  With  numerous  Ex- 
amples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC  MECHANICS. 

With  numerous  Examples. 
AN    ELEMENTARY    TREATISE    ON     HYDROMECHANICS. 

With  numerous  Examples. 


COLLEGE  ALGEBRA, 


FOR  THE  USE  OF 


ACADEMIES,  COLLEGES,  AND  SCIENTIFIC 
SCHOOLS. 


WITBE  NUMEROUS  EXAMPLES. 


BY 


EDWARD  A.  BOWSER,  LL.D., 

PROFESSOR   OF   MATHEMATICS   AND   ENGINEERING    IN    RUTGERS    COLLEGE, 


FOURTH  EDITION. 


BOSTON,   U.S.A.: 
D.   C.   HEATH  &   CO.,   PUBLISHERS. 

1893. 


Copyright.  1888. 


Copyright,  1888, 
By  E.  A.  BOWSER. 


STovtooofc  }0rcBB: 
Berwick  &  Smith,  Boston,  U.S.A. 


t-  _r   — 


PREFACE. 


The  present  work  is  designed  as  a  text-book  for  Acade- 
mies, Colleges,  and  Scientific  Schools.  The  book  is  com- 
plete in  itself ;  it  begins  at  the  beginning  of  the  subject ;  and 
the  full  treatment  of  the  earlier  parts  renders  it  unnecessary 
that  students  who  use  it  shall  have  previously  studied  a  more 
elementary  Algebra. 

The  aim  has  been  to  explain  the  principles  in  as  concise 
and  simple  a  manner  as  was  consistent  with  clearness  and 
careful  accuracy,  and  to  discuss  all  the  elementary  parts  of 
the  subject  as  completely  as  possible  within  the  limits  of  a 
single  volume.  Copious  illustrations  have  been  given  to 
make  the  work  intelligible  and  interesting  to  young  students  ; 
and  numerous  explanatory  notes  have  been  all  along  inserted, 
to  guard  the  pupil  against  the  errors  to  which  experience 
shows  he  is  liable.  Sturm's  theorem  has  been  omitted, 
because  it  has  no  application  in  the  usual  course  of  mathe- 
matical study,  and  because  the  mental  power  of  the  student 
can  be  spent  more  profitably  on  other  branches  of  mathe- 
matics. 

In  the  earlier  chapters,  some  of  the  most  interesting 
practical  applications  of  the  subject  have  been  introduced. 
Thus,  a  chapter  on  easy  equations  and  problems  precedes 
the  chapters  on  Factoring  and  Fractions.  By  this  course  the 
beginner  soon  becomes  acquainted  with  the  ordinary  Alge- 
braic processes  without  encountering  too  many  difficulties  ; 
and  he  is  at  the  same  time  deriving  the  pleasure  which  a 

797947 


VI  PREFACE. 

student  always  feels  in  using  his  knowledge  to  some  practical 
end. 

Throughout  the  book  are  numerous  examples  fully  worked 
out,  to  illustrate  the  most  useful  applications  of  important 
rules,  and  to  exhibit  the  best  methods  of  arranging  the  work. 
No  principle  is  well  learned  by  a  pupil  and  thoroughly  iixed 
in  his  mind  till  he  can  use  it.  For  this  purpose  a  very  large 
number  of  examples  is  given  at  the  ends  of  the  chapters. 
These  examples  have  been  selected  and  arranged  so  as  to 
illustrate  and  enforce  every  part  of  the  subject.  Each  set 
has  been  carefully  graded,  commencing  with  some  which  are 
very  easy,  and  proceeding  to  others  which  are  more  difficult. 
Complicated  examples  have  been  excluded,  because  they 
consume  time  and  energy  which  may  be  otherwise  spent 
more  profitably.  The  numerous  examples  are  given  for  the 
convenience  of  the  teacher,  that  he  may  have,  year  by  year, 
in  using  the  book,  a  sufficient  variety  from  which  to  select, 
and  so  avoid  routine,  rather  than  to  require  any  one  pupil 
to  do  them  all. 

In  preparing  this  work  I  have  consulted  the  writings  of 
some  of  the  best  authors.  The  chief  sources  from  which 
I  have  derived  assistance  are  the  Treatises  of  Wood, 
De  Morgan,  Serret,  Hind,  Young,  Todhunter,  Coleuzo, 
Hall  and  Knight,  Smith,  Chrystal,  and  Whitworth. 

It  remains  for  me  to  express  my  thanks  to  those  of  my 
friends  who  have  kindly  assisted  me  in  reading  and  correct- 
ing the  proof-sheets  and  verifying  copy.  My  especial  thanks 
are  due  to  my  old  pupil,  Prof.  R.  W.  Prentiss  of  the 
Nautical  Almanac  Office,  for  reading  the  MS.,  and  for 
valuable  suggestions. 

E.    A.    B. 
Rutgers  College, 

New  Brunswick,  N.J.,  May,  1888. 


TABLE    OF    CONTENTS. 


CHAPTER  I. 

FIRST     PRINCIPLES. 

ART.  PAGE 

1.  Quantity  and  its  Measure 1 

2.  Number 1 

3.  Mathematics 2 

4.  Algebra 2 

5.  Algebraic  Symbols 2 

6.  Symbols  of  Quantity 2 

7.  Symbols  of  Operation 3 

8.  The  Sign  of  Addition 3 

9.  The  Sign  of  Subtraction 3 

10.  The  Sign  of  Multiplication 4 

11.  The  Sign  of  Division 5 

12.  The  Exponential  Sign 6 

13.  The  Radical  Sign 7 

14.  Symbols  of  Relation 7 

15.  Symbols  of  Abbreviation 8 

16.  Algebraic  Expressions 9 

17.  Factor  — Coefficient 11 

18.  A  Term,  its  Dimensions,  and  Degree  —  Homogeneous  ...  12 

19.  Simple  and  Compound  Expressions 13 

20.  Positive  and  Negative  Quantities 13 

21.  Additions  and  Multiplications  Made  in  any  Order    ....  17 

22.  Suggestions  for  the  Student  in  Solving  Examples     ....  18 
Examples 18 

CHAPTER  II. 

ADDITION. 

23.  Addition  — Algebraic  Sum ■ 21 

24.  To  Add  Terms  which  are  Like  and  have  Like  Signs     ...  21 

25.  To  Add  Terms  which  are  Like,  but  have  Unlike  Signs     .     .  22 

26.  To  Add  Terms  which  are  not  all  Like  Terms 22 

27.  Remarks  on  Addition 23 

Examples 24 

vii 


viii  CONTENTS. 


CHAPTER  III. 

SUBTRACTION. 

ART.  PAGB 

28.  Subtraction  —  Algebraic  Difference 26 

29.  Rule  for  Algebraic  Subtraction 26 

30.  Remarks  on  Addition  and  Subtraction 29 

31.  The  Use  of  Parentheses 30 

32.  Plus  Sign  before  the  Parenthesis 31 

33.  Minus  Sign  before  the  Parenthesis 31 

34.  Compound  Parentheses 32 

Examples 33 


CHAPTER  IV. 

MULTIPLICATION. 

35.  Multiplication 38 

36.  Rule  of  Signs 38 

37.  The  Multiplication  of  Monomials 41 

38.  To  Multiply  a  Polynomial  by  a  Monomial 42 

39.  Multiplication  of  a  Polynomial  by  a  Polynomial 43 

40.  Multiplication  by  Inspection 48 

41.  Special  Forms  of  Multiplication  —  Formulae 49 

42.  Important  Results  in  Multiplication 53 

43.  Results  of  Multiplying  Algebraic  Expressions 54 

Examples 55 


CHAPTER  V. 

DIVISION. 

44.  Division 68 

4.").  The  Division  of  one  Monomial  by  Another 58 

46.  Tbe  Rule  of  Signs 60 

47.  To  Divide  a  Polynomial  by  a  Monomial 61 

48.  To  Divide  one  Polynomial  by  Another 61 

49.  Division  with  the  Aid  of  Parentheses 67 

50.  Where  the  Division  cannot  he  Exactly  Performed    ....  67 

51.  Important  Examples  in  Division 68 

Examples 69 


CONTENTS. 


CHAPTER   VI. 

SIMPLE    EQUATIONS    OF    ONE    UNKNOWN    QUANTITY. 

iRT.  PAGE 

52.  Equations  —  Identical  Equations 72 

53.  Equation  of  Condition  —  Unknown  Quantity 72 

54.  Axioms 73 

55.  Clearing  of  Fractions 74 

56.  Transposition 75 

57.  Solution  of  Simple  Equations  with  one  Unknown  Quantity  .  76 

58.  Fractional  Equations 79 

59.  To  Solve  Equations  whose  Coefficients  are  Decimals    ...  81 

60.  Literal  Equations 81 

61.  Problems  Leading  to  Simple  Equations 83 

Examples 89 


CHAPTER  VII. 

FACTORING  —  GREATEST   COMMON   DIVISOR — LEAST   COMMON 
MULTIPLE. 

62.  Definitions 96 

63.  When  All  the  Terms  Have  one  Common  Factor 96 

64.  Expressions  Containing  Four  Terms 97 

65.  To  Factor  a  Trinomial  of  the  Form  x2  +  ax  +  b     ....    99 

66.  To  Factor  a  Trinomial  of  the  Form  ax'1  +  bx  +  c    .     .     .     .101 

67.  To  Factor  the  Difference  of  Two  Squares 104 

68.  When  One  or  Both  of  the  Squares  is  a  Compound  Expression    105 

69.  Compound  Quantities  as  the  Difference  of  Two  Squares  .     .  106 

70.  To  Factor  the  Sum  or  the  Difference  of  Two  Cubes     .     .     .  107 

71.  Miscellaneous  Cases  of  Resolution  into  Factors 107 

Examples 108 

GREATEST    COMMON    DIVISOR. 

72.  Definitions Ill 

73.  Monomials,  and  Polynomials  which  can  be  Easily  Factored  .  Ill 

74.  Expressions  which  cannot  be  Easily  Factored 113 

Examples 115 


CONTENTS. 


LEAST    COMMON    MULTIPLE. 

TAGE 


ART. 

75.  Definitions 120 

76.  Monomials,  and  Polynomials  which  can  be  Easily  Factored  .  li'l 

77.  Expressions  which  cannot  be  Easily  Factored 123 

Examples 125 


CHAPTER   VIII. 

FRACTIONS. 

78.  A  Fraction  — Entire  and  Mixed  Quantities 128 

79.  To  Reduce  a  Fraction  to  its  Lowest  Terms 129 

80.  To  Reduce  a  Mixed  Quantity  to  the  Form  of  a  Fraction    .     .  132 

81.  To  Reduce  a  Fraction  to  an  Entire  or  Mixed  Quantity .     .     .  i:>5 

82.  To  Reduce  Fractions  to  their  Least  Common  Denominator  .  134 

83.  Rule  of  Signs  in  Fractions 136 

84.  Addition  and  Subtraction  of  Fractions 138 

85.  To  Multiply  a  Fraction  by  an  Integer 144 

86.  To  Divide  a  Fraction  by  an  Integer 144 

87.  To  Multiply  Fractions ]47> 

88.  To  Divide  Fractions 147 

89.  Complex  Fractions 148 

90.  A  Single  Fraction  Expressed  as  a  Group  of  Fractions  .     .     .151 
Examples lu2 


CHAPTER  IX. 

HARDER   SIMPLE   EQUATIONS   OF   ONE   UNKNOWN   QUANTITY. 

91.  Solution  of  Harder  Equations 157 

92.  Harder  Problems  Leading  to  Simple  Equations 161 

Examples *^ 

CHAPTER  X. 

SIMULTANEOUS   SIMPLE  EQUATIONS   OF  TWO   OR   MORE   UNKNOWN 
QUANTITIES. 

08.  Simultaneous  Equations  of  Two  Unknown  Quantities     .    .  179 

L81 

181 


94.  Elimination .    . 

95.  Elimination  by  Addition  or  Subtraction 


CONTENTS.  XI 

ART.  PAGE 

96.  Elimination  by  Substitution 185 

97.  Elimination  by  Comparison , 186 

98.  Fractional  Simultaneous  Equations 188 

99.  Literal  Simultaneous  Equations 189 

100.  Simultaneous  Equations  witb  Tbree  Unknown  Quantities    .  191 

101.  Problems  Leading  to  Simultaneous  Equations 196 

Examples 202 


CHAPTER  XI. 

INDETERMINATE    PROBLEMS  —  DISCUSSION   OP   PROBLEMS 

INEQUALITIES. 

102.  Indeterminate  Equations  —  Impossible  Problems     .     .    .     .211 

103.  Discussion  of  Problems  —  Negative  Results 213 

104.  Interpretation  of  the  Forms,  --,  — ,  £ 216 

0     oo     0 

105.  Problem  of  the  Couriers 218 

INEQUALITIES. 

106.  Inequalities 221 

Examples 225 


CHAPTER  XII. 

INVOLUTION   AND   EVOLUTION. 

107.  Involution '  . 227 

108.  Involution  of  Powers  of  Monomials 227 

109.  Involution  of  Binomials 229 

110.  Involution  of  Polynomials 230 

EVOLUTION. 

111.  Evolution  — Evolution  of  Monomials 232 

112.  Square  Root  of  a  Polynomial 234 

113.  Square  Root  of  Arithmetic  Numbers 238 

114.  Square  Root  of  a  Decimal 240 

115.  Cube  Root  of  a  Polynomial 243 

116.  Cube  Root  of  Aritbmetic  Numbers 246 

116a.  Cube  Root  of  a  Decimal 24S 

Examples 250 


Xii  CONTENTS. 


CHAPTER   XIII. 

THE  THEORY  OF  EXPONENTS  —  SURDS. 

ABT.  TAGS 

117.  Exponents  that  are  Positive  Integers 253 

US.  Fractional  Exponents 25? 

119.  Negative  Exponents 255 

120.  To  Prove  that  (am)»  =  amn  is  True  for  All  Values  of  m  and  n    257 

121.  To  Prove  that  (ab)n  =  anbn  for  Any  Value  of  n 258 

SURDS     (RADICALS). 

122.  Surds  — Definitions 2G1 

123.  To  Reduce  a  Rational  Quantity  to  a  Surd  Form 262 

124.  To  Introduce  the  Coefficient  Under  the  Radical  Sign    .     .     .  262 

125.  To  Reduce  an  Entire  to  a  Mixed  Surd 262 

126.  Reduction  of  Surds  to  Equivalent  Surds 263 

127.  Addition  and  Subtraction  of  Surds 2<>4 

128.  Multiplication  of  Surds 265 

129.  To  Rationalize  the  Denominator  of  a  Fraction 267 

130.  Division  of  Surds 268 

131.  Binomial  Surds  —  Important  Propositions 269 

132.  Square  Root  of  a  Binomial  Surd 270 

133.  Equations  Involving  Surds 272 

Examples 273 


CHAPTER   XIV. 

QUADRATIC    EQUATIONS    OP   ONE   UNKNOWN   QUANTITY. 

134.  Quadratic  Equations 280 

135.  Pure  Quadratic  Equations 280 

136.  Affected  Quadratic  Equations 282 

137.  Condition  for  Equal  Roots 286 

138.  Hindoo  Method  of  Completing  the  Square 2S8 

139.  Solving  a  Quadratic  by  Factoring 290 

140.  To  Form  a  Quadratic  when  the  Roots  are  Given i'!»2 

141.  Equations  Having  Imaginary  Roots 284 

142.  Equations  of  Higher  Degree  than  the  Second 295 

143.  Solutions  by  Factoring 298 

144.  Problems  Leading  to  Quadratic  Equations 300 

Examples 808 


CONTENTS. 

CHAPTER  XV. 

SIMULTANEOUS    QUADRATIC    EQUATIONS. 


ART 


316 
318 


PAGE 

145.  Simultaneous  Quadratic  Equations 311 

146.  When  One  of  the  Equations  is  of  the  First  Degree  .     .     .     .311 

147.  Equations  of  the  Form  x  ±  y  =  a,  and  xy  =  b 312 

148.  When  the  Equations  Contain  a  Common  Algebraic  Factor    .  314 

149.  Homogeneous  Equations  of  the  Second  Degree    .    .     . 

150.  When  the  Two  Equations  are  Symmetrical 

151.  Special  Methods 319 

152.  Quadratic  Equations  with  Three  Unknown  Quantities      .     .  321 

153.  Problems  Leading  to  Simultaneous  Quadratic  Equations  .     .  321 
Examples 324 


CHAPTER    XVI. 

RATIO PROPORTION VARIATION. 

154.  Ratio  —  Definitions 331 

155.  Properties  of  Ratios 333 

PROPORTION. 

156.  Definitions 336 

157.  Properties  of  Proportions 337 

VARIATION. 

158.  Definition 342 

159.  Different  Cases  ot  Variation 343 

160.  Propositions  in  Variation 344 

Examples 348 


CHAPTER  XVII. 

ARITHMETIC,    GEOMETRIC,    AND    HARMONIC   PROGRESSIONS. 
ARITHMETIC    PROGRESSION. 

161.  Definitions  —  Formulae 353 

162.  Arithmetic  Mean 356 


xiv  ,        CONTENTS. 

GEOMETRIC   PROGRESSION. 

ART.  PAGE 

163.  Definition  —  Formulae 359 

164.  Geometric  Mean 361 

165.  The  Sum  of  an  Infinite  Number  of  Terms ,>  362 

166.  Value  of  a  Repeating  Decimal 364 

HARMONIC    PROGRESSION. 

167.  Definition 365 

168.  Harmonic  Mean 366 

169.  Relation  between  the  Different  Means 368 

Examples 369 


CHAPTER  XVIII. 

MATHEMATICAL   INDUCTION  —  CONTINUED    FRACTIONS  — 
PERMUTATIONS    AND    COMBINATIONS. 

170.  Mathematical  Induction  —  General  Proofs  of  Theorems  .    .  374 

CONTINUED    FRACTIONS. 

171.  Definition 376 

172.  To  Convert  a  Given  Fraction  into  a  Continued  Fraction  .     .  376 

173.  The  Convergents  are  Alternately  too  Great  and  too  Small    .  377 
•  174.  Law  of  Formation  of  the  Successive  Convergents    ....  378 

175.  Difference  between  Two  Consecutive  Convergents    ....  379 

176.  Limit  of  Error  in  Taking  any  Convergent 380 

177.  Approximate  Value  of  a  Quadratic  Surd 383 

178.  Periodic  Continued  Fractions 384 

PERMUTATIONS   AND   COMBINATIONS. 

179.  Definitions 385 

180.  The  Number  of  Permutations 386 

181.  The  Number  of  Combinations 388 

182.  To  Divide  m  +  n  Things  into  Two  Classes 800 

183.  Permutations  of  ?i  Things  not  all  Different 801 

Examples 392 


CONTENTS. 

CHAPTER   XIX. 

INDETERMINATE    COEFFICIENTS PARTIAL    FRACTIONS  - 

BINOMIAL    THEOREM. 


ART. 


PAOK 


184.  Definitions oWj 

185.  Indeterminate  Coefficients 396 

186.  Development  by  Indeterminate  Coefficients 397 

PARTIAL    FRACTIONS. 

187.  Partial  Fractions.     Case  1 401 

188.  Partial  Fractions.     Case  II .402 

189.  Partial  Fractions.     Case  III 403 

BINOMIAL   THEOREM. 

190.  Positive  Integral  Exponent 405 

191.  The  rth  or  General  Term  of  the  Expansion .......  408 

192.  Any  Exponent 409 

Examples 413 


CHAPTER  XX. 

SUMMATION     OF     SERIES. 
RECURRING   SERIES. 

193.  Definition 416 

194.  The  Scale  of  Relation  being  Given,  to  Find  Any  Term     .     .  417 

195.  To  Find  the  Sum  of  n  Terms  of  a  Recurring  Series  .     .     .     .418 

196.  Series  Summed  by  Means  of  Other  Series 419 

METHOD    OF    DIFFERENCES. 

197.  Definition  — To  Find  Any  Term  of  a  Series 421 

198.  To  Find  the  Sum  of  n  Terms  of  Any  Series     .  ...  422 

199.  Piles  of  Cannon-Bails .424 

200.  Interpolation 426 

201.  Reversion  of  Series 427 

Examples 428 


CONTENTS. 


CHAPTER    XXI. 

LOGARITHMS EXPONENTIAL   AND    LOGARITHMIC   SERIES 

INTEREST    AND    ANNUITIES. 

LOGARITHMS. 
ART.  TAGE 

202.  Definitions 433 

203.  Properties  of  Logarithms 434 

204.  Comparison  of  Two  Systems  of  Logarithms 430 

205.  Common  System  of  Logarithms 437 

206.  Table  of  Logarithms 439 

207.  Exponential  Equations 442 

EXPONENTIAL   AND    LOGARITHMIC   SERIES. 

208.  Exponential  Series 443 

209.  Logarithmic  Series 444 

210.  Computation  of  Logarithms 445 

INTEREST   AND    ANNUITIES. 

211.  Simple  Interest 447 

212.  Present  Value  and  Discount .     .         448 

213.  Compound  Interest 449 

214.  Present  Value  and  Discount 450 

215.  Annuities 452 

216.  The  Amount  of  an  Annuity 452 

217.  The  Present  Value  of  an  Annuity 453 

218.  The  Present  Value  of  a  Deferred  Annuity 454 

Examples 450 

CHAPTER  XXII. 

IMAGINARY    QUANTITIES INDETERMINATE    EQUATIONS 

THEORY    OF    NUMBERS. 

IMAGINARY    QUANTITIES. 

219.  Definitions 460 

220.  Fundamental  Principles 461 

221.  Geometric  Meaning  of  the  Imaginary  Unit 462 

INDETERMINATE    EQUATIONS. 

222.  Indeterminate  Equations t,;' 


CONTENTS. 


THEORY    OF    NUMBERS. 
ART.  PAGE 

223.  Scales  of  Notation 468 

224.  To  Express  a  Whole  Number  in  Any  Proposed  Scale    .     .     .  470 

225.  Properties  of  Numbers  in  the  Common  System 472 

226.  Theorems  in  Relation  to  Numbers 474 

Examples 478 


CHAPTER   XXIII. 

PROBABILITY    (CHANCE). 

227.  Definition 481 

228.  Simple  Events 483 

229.  Compound  Events 485 

230.  Mutually  Exclusive  Events 488 

231.  Expectation 492 

232.  General  Problem 492 

Examples 494 


CHAPTER   XXIV. 

THEORY     OF     EQUATIONS. 

233.  General  Equation  of  the  nth  Degree 497 

234.  Divisibility  of  Equations 498 

235.  Number  of  Roots 499 

236.  Relations  between  Coefficients  and  Roots 500 

237.  Fractional  Roots 502 

238.  Imaginary  Roots 502 

239.  Descartes'  Rule  of  Signs 503 

240.  Derived  Functions 505 

241.  Equal  Roots 507 

TRANSFORMATION   OF   EQUATIONS. 

242.  Equation  with  Roots  of  Sign  opposite  those  of  f{x)  =  0   .     .  509 

243.  Equation  with  Roots  Multiples  of  those  of  f{x)  =0      ...  510 

244.  Equation  with  Roots  Reciprocals  of  those  of  f(x)  -  0  .     .     .511 

245.  Equation  with  Roots  Squares  of  those  of  f{x)  =  0   .     .     .     .511 

246.  Equation  with  Roots  Less  than  those  of  f(x)  =  0     .     .    .    .  512 

247.  Horner's  Method  of  Synthetic  Division 513 

248.  Removal  of  Any  Given  Term 515 


xviii  CONTENTS. 


LIMITS    OF    THE    REAL    HOOTS    OF    AN    EQUATION. 
ART.  PAGE 

249.  Definition 515 

250.  Superior  Limit 51(5 

251.  Inferior  Limit 517 

252.  Limits  between  which  the  Roots  separately  Lie 518 

Examples 519 


CHAPTER    XXV. 

SOLUTION    OF   HIGHER   NUMERICAL    EQUATIONS. 

253.  Commensurable  Roots 523 

254.  Reciprocal  Equations 525 

255.  Binomial  Equations 527 

256.  Cardan's  Solution  of  a  Cubic  Equation 529 

257.  Incommensurable  Roots 532 

258.  Horner's  Method  of  Approximation 532 

259.  Newton's  Method  of  Approximation 536 

260.  Approximation  by  Double  Positiou 537 

Examples 539 


ALGEBRA 


CHAPTER     I. 

FIRST     PRINCIPLES. 

t 

1.  Quantity  and  its  Measure.  —  Quantity  is  any 
thing  that  is  capable  of  increase,  diminution,  and  measure- 
ment ;  as  time,  space,  motion,  weight,  and  area. 

To  measure  a  quantity  is  to  find  how  many  times  it  con- 
tains another  quantity  of  the  same  kind,  taken  as  a  standard 
of  comparison.     This  standard  is  called  a  unit. 

For  example,  if  we  wish  to  determine  the  quantity  of  a  weight,  we 
must  take  a  unit  of  weight,  such  as  a  pound,  or  an  ounce,  and  observe 
how  many  times  it  is  contained  in  the  quantity  to  be  measured.  If 
we  wish  to  measure  area,  we  must  take  a  unit  of  area,  as  a  square 
foot,  square  yard,  or  acre,  and  see  how  many  times  it  is  contained  in 
the  area  to  be  measured.  So  also,  if  we  wish  to  measure  the  value  of 
a  sum  of  money,  or  any  portion  of  time,  we  must  take  a  unit  of  value, 
as  a  dollar  or  a  sovereign,  or  a  unit  of  time,  as  a  day  or  a  year,  and  see 
how  many  times  it  is  contained  in  the  quantity  to  be  measured. 

2.  Number.  —  The  relation  between  any  quantity  and  its 
unit  is  always  expressed  by  a  number;  a  number  therefore 
simply  shows  hoio  many  times  any  quantity  to  be  measured 
contains  another  quantity,  arbitrarily  assumed  as  the  unit. 
All  quantities,  therefore,  can  be  expressed  by  numbers. 

All  numbers  are  concrete  or  abstract. 

A  Concrete  Number  is  one  in  which  the  kind  of  quantity 
which  it  measures  is  expressed  or  understood ;  as  6  books, 
10  men,  4  days. 

1 


2  MATH  EM  A  TICS.  —  ALGEBRA. 

An  Abstract  Number  is  one  in  which  the  kind  of  quantity 
which  it  measures  is  not  expressed  ;  as  6,  10,  4. 

The  word  quantity  is  often  used  with  the  same  meaning  as  number. 
Numbers  may  be  either  whole  or  fractional.  The  word  integer  is  often 
used  instead  of  whole  number. 

3.  Mathematics.  —  Mathematics  is  the  science  which 
treats  of  the  measurement  and  relations  of  quantities.  It 
is  divided  into  two  parts,  Pure  Mathematics  and  Mixed 
Mathematics. 

Pure  Mathematics  consists  of  the  four  branches,  Arithmetic, 
Algebra,  Geometry,  and  Calculus. 

Mixed  Mathematics  is  the  application  of  Pure  Mathematics 
to  the  Mechanic  Arts. 

4.  Algebra.  —  Algebra  is  that  branch  of  Mathematics  in. 
which  we  reason  about  numbers  by  means  of  symbols.  The 
different  symbols  used  represent  the  numbers  themselves, 
the  manner  in  which  they  are  related  to  one  another,  and  the 
operations  performed  on  them. 

In  Arithmetic,  numbers  are  represented  by  ten  characters,  called 
figures,  which  are  variously  combined  according  to  certain  rules,  and 
which  have  but  one  single  definite  value.  In  Algebra,  on  the  contrary, 
numbers  are  represented  either  by  figures,  as  in  Arithmetic,  or  by 
symbols  which  may  have  any  value  we  choose  to  assign  to  them. 

5.  Algebraic  Symbols.  —  The  symbols  employed  in 
Algebra  are  of  four  kinds  :  symbols  of  quantity,  symbols  of 
operation,  symbols  of  relation,  and  symbols  of  abbreviation. 

6.  Symbols  of  Quantity.  —  The  symbols  of  quantity 
may  be  any  characters  whatever,  but  those-  that  are  most 
commonly  used  are  figures  and  the  letters  of  the  alphabet; 
and  as  in  the  simplest  mathematical  problems  there  aid 
certain  quantities  given,  in  order  to  determine  other  quail' 
tities  which  are  unknown,  it  is  usual  to  represent  the  known 
quantities  by  figures  and  by  the  first  letters  of  the  alphabet, 
a,  b,  c,  etc.  ;  a',  b' ,  c',  etc.,  read  a  prime,  b  prime,  c  prime% 
etc. ;  av  bv  cv  etc.,  read  a  one,  b  one,  c  one,  etc. ;  while  the 


SYMBOLS   OF   OPERATION.  3 

unknown  quantities  are  represented  by  the  final  letters  of 
the  alphabet,  v,  x,  y,  z,  i/,  a/,  y',  z',  etc. 

Known  Quantities  are  those  whose  values  are  given. 

Unknown  Quantities  are  those  whose  values  are  required. 

Since  all  quantities  can  be  expressed  by  numbers  (Art.  2), 
it  is  only  these  numbers  with  which  we  are  concerned,  and 
the  symbols  of  quantity,  whether  figures  or  letters,  always 
represent  members. 

In  Arithmetic  a  character  has  but  one  definite  and  invariable  value, 
while  in  Algebra  a  symbol  may  stand  for  any  quantity  we  choose  to 
assign  to  it  (Art.  4);  but  while  there  is  no  restriction  as  to  the 
numerical  values  a  symbol  may  represent,  it  is  understood  that  in  the 
same  piece  of  work  it  keeps  the  same  value  throughout.  Thus,  when 
we  say  "let  a  —  2,"  we  do  not  mean  that  a  must  have  the  value  2 
always,  but  only  in  the  particular  example  we  are  considering.  Also, 
we  may  operate  with  symbols  without  assigning  to  them  any  particular 
value  at  all;  and  it  is  with  such  operations  that  Algebra  is  chiefly 
concerned. 

7.  Symbols  of  Operation.  —  The  symbols  of  operation 
are  the  same  in  Algebra  as  in  Arithmetic,  or  in  any  other 
branch  of  Mathematics,  and  are  the  following : 

8.  The  Sign  of  Addition,  +,  is  called  plus.  When 
placed  before  a  number  it  denotes  that  the  number  is  to 
be  added.  Thus,  6+3,  read  6  phis  3,  means  that  3  is  to  be 
added  to  6  ;  a  +  b,  read  a  pZws  6,  denotes  that  the  number 
represented  by  b  is  to  be  added  to  the  number  represented 
by  a ;  or,  more  briefly,  it  denotes  that  b  is  to  be  added  to  a. 
If  a  represent  8,  and  b  represent  5,  then  a  +  b  represents  13. 

Similarly  a  +  b  +  c,  read  a  plus  b  plus  c,  denotes  that 
we  are  to  add  b  to  a,  and  then  add  c  to  the  result. 

9.  The  Sign  of  Subtraction,  — ,  is  called  minus.  When 
placed  before  a  number  it  denotes  that  the  number  is  to  be 
subtracted.  Thus,  a  —  b,  read  a  minus  b,  denotes  that  the 
number  represented  by  b  is  to  be  subtracted  from  the  number 
represented  by  a  ;  or,  more  briefly,  that  b  is  to  be  subtracted 
from  a.  If  a  represent  8,  and  b  represent  5,  then  a  —  b 
represents  3. 


4  THE  SIGN   OF  MULTIPLICATION. 

Similarly  a  —  b  —  c,  read  a  minus  b  minus  c,  denotes 
that  we  are  to  subtract  b  from  a,  and  then  subtract  c  from 
the  result. 

If  neither  +  nor  —  stands  before  a  quantity,  -f-  is  always 
understood  ;  thus  a  means  +  a. 

Quantities  which  have  the  same  sign,  either  +  or  — ,  are 
said  to  have  like  signs.  Thus,  +  a  and  -f-  b  have  like  signs, 
also  —  a  and  —  b  ;  but  -f  a  and  —  b  have  unlike  signs. 

Note.  —  Although  there  are  many  signs  used  in  Algebra,  when  the 
sign  of  a  quantity  is  spoken  of,  it  means  the  +  or  —  sign  which  is 
prefixed  to  it;  and  when  we  speak  of  changing  the  signs  of  an  expres- 
sion, it  means  that  we  are  to  change  +  to  —  and  —  to  +  wherever 
they  occur. 

The  sign  ~  is  sometimes  used  to  denote  the  difference  of  two 
numbers  when  it  is  not  known  which  of  them  is  the  greater.  Thus, 
a  ~  b  denotes  the  difference  of  the  numbers  represented  by  a  and  b; 
and  is  equal  to  a  —  b,orb  —  a,  according  as  a  is  greater  or  less  than  b; 
but  this  symbol  ~  is  very  rarely  required. 

10.  The  Sign  of  Multiplication,  x ,  is  read  into,  or 
times,  or  multiplied  by.  When  placed  between  two  numbers 
it  denotes  that  they  are  to  be  multiplied  together.  Thus, 
a  x  b,  read  a  into  b,  denotes  that  the  number  represented  by 
a  is  to  be  multiplied  by  the  number  represented  by  b,  or, 
more  briefly,  that  a  is  to  be  multiplied  by  b,  or  that  the  two 
are  to  be  multiplied  together.  The  numbers  to  be  multiplied 
together  are  called  factors,  and  the  result  of  the  multiplica- 
tion is  called  a  product.  Thus  5,  a,  and  b  are  the  factors 
of  the  product  5  X  a  X  6.  If  a  represent  8,  and  b  repre- 
sent 4,  then  a  X  b  represents  32  ;  a  and  b  are  the  factors  of 
the  product  a  X  b,  or  8  and  4  are  the  factors  of  32.  Simi- 
larly a  X  b  X  c  denotes  the  product  of  the  numbers  a,  6, 
and  c.  If  a  represent  G,  b  represent  8,  and  c  represent  10, 
then  a  x  b  x  c  represents  480,  and  5  x  a  x  b  x  c  repre- 
sents 2400. 

Sometimes  a  point  is  used  instead  of  the  sign  x  ;  or.  still 
mure  commonly,  one   number  is  placed  close  after  the  other 


THE  SIGN   OF  DIVISION.  5 

without  any  sign  between  them.  Thus,  a  x  b,  a-b,  and  ab 
all  mean  the  same  thing,  viz.,  the  product  of  a  and  b  ;  also, 
a  X  b  X  c,  or  a  ■  b  •  c,  or  abc,  denotes  the  product  of  the 
numbers  a,  6,  and  c.  If  a,  6,  and  c  represent  2,  5,  and  10 
respectively,  then  abc  represents  100. 

If  one  factor  of  a  product  is  equal  to  0,  the  whole  product 
must  be  equal  to  0,  whatever  values  the  other  factors  may 
have.     A  factor  0  is  sometimes  called  a  "zero  factor."  * 

The  sign  of  multiplication  must  not  be  omitted  when  num- 
bers are  expressed  in  the  ordinary  way  by  figures.  Thus  23 
cannot  be  used  to  represent  the  product  of  2  and  3,  because 
23  is  used  to  mean  the  number  twenty-three.  Nor  can  the 
product  of  2  and  3  be  represented  by  2.3,  because  2.3  is 
used  to  mean  two  and  three-tenths.  We  must  therefore 
represent  the  product  of  2  and  3  by  placing  the  sign  of 
multiplication  between  them,  as  follows :  2  x  3.  When  the 
numbers  to  be  multiplied  together  are  represented  by  letters, 
or  by  letters  and  a  figure,  it  is  usual  to  omit  the  sign  of 
multiplication  for  the  sake  of  brevity,  and  write  them  in 
succession  close  to  each  other;  thus,  the  product  of  the 
numbers  7,  a,  6,  c,  and  d  would  be  written  7 abed,  instead  oi 
7  x  a  x  b  x  c  x  d,  or  7  •  a  •  b  •  c  •  d,  and  would  have  the 
same  meaning. 

11.  The  Sign  of  Division,  -5-,  is  read  divided  by,  or 
simply  by.  When  placed  between  two  numbers,  it  denotes 
that  the  number  which  precedes  it  is  to  be  divided  by  the 
number  which  follows  it.  Thus,  a  -s-  6,  read  a  divided  by  b, 
or  a  by  b,  denotes  that  the  number  represented  by  a  is  to  be 
divided  by  the  number  represented  by  6,  or,  more  briefly, 
that  a  is  to  be  divided  by  6.  If  a  represent  8,  and  b  repre- 
sent 2,  then  a  -s-  6  represents  4.  Most  frequently,  to  express 
division,  the  number  to  be  divided  is  placed  over  the  other 


*  It  is  a  common  mistake  of  beginners  to  say  that  an  Algebraic  expression  like 
a  x  0  or  0  x  a  is  equal  to  a,  by  supposing  it  to  mean  a  not  multiplied  at  all ;  whereas 
a  x  0  01  0  x  a  signifieB  0  takeu  a  times,  or  a  taken  0  times,  aud  is  therefore  equal 
toO. 


6  THE  EXPONENTIAL   SIGN. 

with  a  horizontal  line  between  them,  in  the  manner  of   ?i 

fraction  in  Arithmetic.     Thus,  -  is  used  instead  of  a  ■+-  b, 
b 

and  has  the  same  meaning.     Also,  the  sign  of  division  may 

be  replaced  by  a  vertical  line,  straight  or  curved.     Thus, 

a\b,  or  b)a  is  used  instead  of  a  -h  b,   and  has  the  same 

meaning. 

Note.  —  It  is  important  for  the  student  to  notice  the  order  of  the 
operations  in  such  expressions  as  a  -\-  b  x  c  and  a  —  b  -f  c.  The 
former  means  that  b  is  first  to  be  multiplied  by  c,  and  the  result  added 
to  a.  The  latter  means  that  b  is  first  to  be  divided  by  c,  and  the  result 
subtracted  from  a. 

12.  The  Exponential  Sign.  —  This  sign  is  a  small  fig- 
ure or  letter  written  at  the  right  of  and  above  a  number  to 
show  how  many  times  the  number  is  taken  as  a  factor,  and 
is  called  an  exponent.  Thus,  a8  is  used  to  denote  a  x  a,  or 
that  a  is  taken  twice  as  a  factor  ;  a3  is  used  to  denote  a  x  a 
X  a,  or  that  a  is  taken  three  times  as  a  factor ;  a4  is  used  to 
denote  a  x  a  X  «  X  a,  or  that  a  is  taken  four  times  as  a 
factor;  and  an  is  used  to  denote  a  x  a  x  a  x  a,  etc.,  to 
it  factors,  or  that  a  is  taken  n  times  as  a  factor.  Similarly 
a264cd3  is  used  to  denote  aabbbbcddd,  and  7a3cd2  is  used  for 
laaacdd. 

If  a  factor  be  multiplied  b}r  itself  any  number  of  times 
the  product  is  called  a  power  of  that  factor.     Thus, 

a  x  a  is  called  the  second  power  of  a,  and  is  written  «2; 
a  x  a  x  fl  is  called  the  third  power  of  a,  and  is  written  «3; 
a  x  a  x  a  x  u  is  called  the  fourth  power  of  a,  and  is  written  a4; 

and  so  on.  Similarly  aaabbc  is  called  the  product  of  the 
third  power  of  a,  the  second  power  of  b,  and  c,  and  is  written 
u3b'~c. 

The  second  power  of  a,  i.e.,  a2,  is  usually  read  a  to  the 
second  power,  or  a  square.  The  third  power  of  a,  i.e.,  a8, 
is  usually  read  a  to  the  third  power,  or  a  cube.  There  are 
no  such  words  in  use  for  the  higher  powers;  the  fourth 
power  of  «,  i.e.,  a4,    is  usually   read  a   to  the  fourth  potcvr, 


TEE  RADICAL   SIGN —  SYMBOLS   OF  RELATION.  7 

or  briefly,  a  fourth  power ;  and  so  on.  When  the  exponent 
is  unity  it  is  omitted.  Thus  we  do  not  write  a1,  but  simply 
a,  which  is  the  same  as  a1,  and  means  a  to  the  first  power. 

13.  The  Radical  Sign,  V  .  —  A  root  of  a  quantity  is  a 
factor,  which,  multiplied  by  itself  a  certain  number  of  times, 
will  produce  the  given  quantity.  The  square  root  of  a 
quantity  is  that  quantity  whose  square  or  second  power  is 
equal  to  the  given  quantity.  Thus  the  square  root  of  16 
is  4,  because  42  is  equal  to  16  ;  the  square  root  of  a2  is  a,  of 
81  is  9. 

The  square  root  of  a  is  denoted  by  sa,  or  more  simply 

vs. 

Similarly  the  cube,  fourth,  fifth,  etc.,  root  of  any  quantity 
is  that  quantity  whose  third,  fourth,  fifth,  etc.,  power  is 
equal  to  the  given  quantity. 

The  roots  are  denoted  by  the  symbols  \/  ,  V  ,  \/  ,  etc. ; 
thus,  'y/27as  denotes  the  cube  root  of  27a3,  which  is  3a, 
because  3a  to  the  third  power  is  27a3.  Similarly  ^32  is  2. 
The  small  figure  placed  on  the  left  side  of  the  symbol  is 
called  the  index  of  the  root.  Thus  2  is  the  index  of  the 
square  root,  3  of  the  cube  root,  4  of  the  fourth  root,  and 
so  on  ;  the  index,  however,  is  generally  omitted  in  denoting 
the  square  root ;  thus  \a  is  written  instead  of  yja. 

The  symbol  V  is  sometimes  called  the  radical  sign. 
When  this  sign  with  the  proper  index  on  the  left  side  of  it 
is  placed  over  a  quantity  it  denotes  that  some  root  of  the 
quantity  is  to  be  extracted. 

14.  Symbols  of  Relation.  —  The  symbols  of  relation 
are  the  following : 

The  sign  of  equality,  =,  is  read  equals,  or  is  equal  to. 
When  placed  between  two  numbers,  it  denotes  that  they  are 
equal  to  each  other.  Thus  a  =  6,  read  a  equals  b,  or  a  is 
equal  to  b,  denotes  that  the  number  represented  by  a  is  equal 
to  the  number  represented  by  b ;  or,  more  briefly,  that  a 
equals  b.     And   a  +  b  =  c  denotes   that   the   sum   of   the 


8  SYMBOLS   OF  ABBREVIATION. 

numbers  a  and  b  is  equal  to  the   number  c ;    so   that   if   a 
represent  8  and  b  represent  4,  then  c  must  represent  12. 

The  signs  of  inequality,  >  and  <,  are  read  is  greater  than, 
and  is  less  than,  respectively.  When  either  is  placed  between 
two  numbers  it  denotes  that  they  are  unequal  to  each 
other,  the  opening  of  the  angle  in  both  cases  being  turned 
towards  the  greater  number.  Thus  a  >  b,  read  a  is  greater 
than  b,  denotes  that  the  number  a  is  greater  than  the  number 
b,  and  b  <  a,  read  b  is  less  than  a,  denotes  that  the  number  b 
is  less  than  the  number  a. 

The  sign  of  ratio,  : ,  is  read  is  to  or  to.  When  placed 
between  two  numbers  it  denotes  their  ratio.  Thus  a  :  b, 
read  a  is  to  b,  or  the  ratio  of  a  to  b,  denotes  the  ratio  of  the 
number  a  to  the  number  b.  A  proportion,  or  two  equal 
ratios,  is  expressed  by  writing  the  sign  =  or  the  sign  :  : 
between  two  equal  ratios.     Thus 

a :  b  =  c  :  d,  or  a  :  b  :  :  c  :  d , 
read  a  is  to  b  as  c  is  to  d,  or  the  ratio  of  a  to  b  equals  the 
ratio  of  c  to  d. 

The  sign  of  variation,  cc ,  is  read  varies  as.  When  placed 
between  two  numbers  it  denotes  that  they  increase  and 
decrease  together,  in  the  same  ratio.  Thus  x  a  ?/,  read  x 
varies  as  y,  denotes  that  x  and  y  increase  and  decrease 
together. 

15.  Symbols  of  Abbreviation.  —  The  symbols  of  abbre- 
viation are  the  following : 

The  signs  of  deduction,  .•.  is  read  hence  or  therefore, 
and  •.•  is  read  .since  or  because. 

The  signs  of  aggregation  are  the  bar  \  ,  the  parenthesis  (  ) , 
the  bracket  [  ] ,  the  brace  \  \ ,  and  the  vinculum  .  These 
are  employed  to  connect  two  or  more  numbers  which  are  to 
be  treated  as  if  they  formed  one  number.  Thus,  suppose 
we  have  to  denote  that  the  sum  of  a  and  6  is  to  be  multiplied 
bye;  we  denote  it  thus  («  +  b)  x  c  or  \a  -f  b\  X  c,  or 
simply  (a  +  b)  c  or  {a  4-  l>\  C,  here  we  mean  that  the 
whole  of  a  +  b  is  to  be  multiplied  by  c.       If  we  omit  the 


ALGEBRAIC  EXPRESSIONS.  9 

parenthesis,  or  brace,  we  have  a  +  be,  and  this  denotes  that 
b  only  is  to  be  multiplied  by  c  and  the  result  added  to  a. 

Also  (a  +  b  +  c)  x  (d  +  e)  denotes  that  the  result 
expressed  by  a  -f-  b  +  c  is  to  be  multiplied  by  the  result  ex- 
pressed by  d  +  e.  This  may  also  be  denoted  simply  thus 
(a  -r-  b  +  c)  (d  +  e),  just  as  a  x  b  is  shortened  into  ab. 
If  we  omit  the  parenthesis  we  have  a  +  b  +  ccZ  +  e,  and 
this  denotes  that  c  only  is  to  be  multiplied  by  d  only,  and  the 
result  added  to  a  '+  b  +  e. 

Also  V  (a  -f-  &  +  c)  denotes  that  we  are  to  obtain  the 
result  expressed  by  a  +  &  +  c,  and  then  take  the  square 
root  of  this  result. 

Also  (ab)2  denotes  ab  X  ab  ;  and  (ab)3  denotes  ab  x  ab  X  a&. 

Also  (a  +  &  +  c)  -=-  (d  +  e)  denotes  that  the  result  ex- 
pressed by  a  +  6  +  c  is  to  be  divided  by  the  result  expressed 
by  d  +  e.  This  may  also  be  expressed  by  the  bracket  thus 
[a  +  b  +  c]  -7-  [d  +  e] ,  or  the  brace  ^a  +'&  +  c\  -=-  |d  +  ej, 

a  +  &   +  c 

or  the  vinculum  a  +  6  +  c-r-d  +  e,  or — 7f~_r~p — >  where 

the  line  between  the  numerator  and  denominator  acts  as  a 
vinculum. 

The  signs  of  continuation  are  dots ,  or  dashes 

,  and  are  read  and  so  on. 

16.  Algebraic  Expressions.  —  The  four  kinds  of  sym- 
bols which  have  been  explained  are  called  Algebraic  symbols 
(Art.  5).  Any  collection  of  Algebraic  symbols  is  called 
an  Algebraic  expression,  or  briefly,  an  expression.  Thus 
4a  +  5b  —  c  +  x  is  an  expression  ;  3&  +  4c  is  the  Alge- 
braic expression  for  3  times  the  number  b  increased  by  4 
times  the  number  c. 

The  numerical  value  of  an  expression  is  the  number 
obtained  by  giving  a  particular  value  to  each  letter,  and 
then  performing  the  operations  indicated. 

We  shall  now  give  some  examples  in  finding  the  numerical 
values  of  expressions,  as  an  exercise  in  the  use  of  the 
symbols  which  have  been  explained. 


10  EXAMPLES. 

EXAMPLES. 

If  a  =  1,  6  =  2,  c  =  3,  a"  =  4,  e  =  5,  find  the  numerical 
values  of  the  following  expressions  : 

1.  9a  +  26  +  3c  —  2d. 

Here  we  have  9a  +  26  +  3c  —  2d  = 

9x1+2x2  +  3x3-2x4  = 
9  +  4  +  9-8  =  14     4n*. 

2.  7ae  +  36c  +  9a\ 

Here  we  have  lae  +  36c  +  9rf  = 

7x1x5  + 3x2x3  +  9x4  =  89   Ans. 

3.  a6ccZ  +  a6ce  +  abde  +  acde  +  6c<2e.  .Ans.  274. 

.     4ac    .    86c       bed  r 

4. 1 .  G. 

6  d  e 

K     cde    ,    bbed       6ade  Q, 

o.    — -  -j — .  a-i. 

ab         ae  be 

If  a  =  1,  6  =  3,  c  =  5,  and  d  =  0,  find  the  numerical 

values  of  the  following  : 

6.  a2  +  262  +  3c2  +  4r/2. 

7.  a4  -  4a86  +  Ga262  -  4a63  +  64. 
g     12a3  -  62  2c2      _  a  +  lr  +  <•' 

3a2  a  +  62  563 

If   a  =  1,   6  =  2,   c  =  3,   d  =  5,   and   e  = 
numerical  values  of   the  following : 
9.    62(«2  +  e2  -  c2). 

10.  ^(26  +  Ad  +  be). 

11.  (a2  +  62  +  c2)(e2  -  d2  -  c2). 

12.  e-\<f(e+\)+2\  +  {e-Te)\IXe-A). 

13.  Find  the  value  of  x2  —  2x  —  9  when  a;  =  5. 
Explanation.  —  If  x  =  any  number,  as  for  example,  5,  then  Xs 

(which  =  X'x)  =  5x,  x3  (which  =  x  -x'2)  =  ox'2,  X*  (which  =  x  .Xs)  =  ox3, 
etc.     Hence  examples  like  18  may  be  solved  as  follows: 


Ans 

.  94. 

16. 

5. 

find 

the 

Ins. 

224. 

8. 

420. 

15. 

x2  -  2x 
x2  =  5x 

—   9  when  x  =  5. 

Explanation. 
x2  =z  5x 

fir  —  2x  =  :'..'• 

3x 

=  15 

C  =  result. 

8x  -  15 

8x  —  9  =  15  —  9 

FACTOR  —  COEFFICIEN  T.  1 1 

14.  Find  the  value  of  x5-  50a;4-  49a3-  100a2-  101a;  -  50, 
when  x  =  51. 

These  examples  may  be  conveniently  solved  as  follows: 
x5  —  50x4  -  49x3  —  lOOe*  —  lOlx  -  50 
+  51      +51      +  102     +  102     +  51 


51 


+      x4  +    2x3  +      2x2  +       x+    1         .-.     result  is  1. 

15.  Find  the  value  of  xi  —  llic3  —  11a;2  —  11a;  —  11  for 
*  =  12.  Ans.  1. 

16.  Find  the  value  of  a;4  -  8a;3  -  19a;2  -  9a;  -  8  for 
x  =  10.  Ans.  3 

17.  Factor  —  Coefficient.  —  When  two  or  more  num- 
bers are  multiplied  together  the  result  is  called  the  product, 
and  each  of  the  numbers  multiplied  together  to  form  the 
product  is  called  a  factor  of  the  product  (Art.  10).  Thus, 
3  X  4  x  5  =  60,  and  each  of  the  numbers  3,  4,  and  5  is  a 
factor  of  the  product  60.  Factors  expressed  by  letters  are 
called  literal  factors ;  factors  expressed  by  figures  are  called 
numerical  factors.  Thus,  in  the  product  Aab,  4  is  called  a 
numerical  factor,  while  a  and  b  are  called  literal  factors. 

The  proof  is  given  in  Arithmetic  that  it  is  immaterial  in 
what  order  the  factors  of  a  product  are  written  ;  it  is  usual, 
however,  to  arrange  them  in  alphabetic  order. 

The  numerical  factor  is  called  the  coefficient  of  the  remain- 
ing factors.  Thus  in  the  expression  Aab,  4  is  the  coefficient, 
and  denotes  that  ab  is  taken  4  times.  But  it  is  sometimes 
convenient  to  consider  any  factor,  or  factors,  of  a  product 
as  the  coefficient  of  the  remaining  factors.  Thus,  in  the 
product  5abc,  5a  may  be  appropriately  called  the  coefficient 
of  be,  or  hob  the  coefficient  of  c. 

The  coefficient  is  called  numerical  or  literal,  according  as  it 
is  a  number,  or  one  or  more  letters.  Thus,  in  the  quantities 
5a;  and  mx,  5  is  a  numerical  and  m  a  literal  coefficient. 

When  no  numerical  coefficient  is  expressed,  1  is  always 
understood.     Thus,  a  is  the  same  as  la. 

A  coefficient  placed  before  any  parenthesis  indicates  that 


12  A    TERM,   ITS  DIMENSIONS,   AND   DEGREE. 

every  term  of  the  expression  within  the  parenthesis  is  to  be 
multiplied  by  that  coefficient. 

Care  must  be  taken  to  distinguish  between  a  coefficient  and 
an  exponent.  Thus  Aa  means  four  times  a,  or  a  +  a  +  a  +  a  ; 
here  4  is  a,  coefficient.  But  a4  means  a  times  a  times  a  times  a, 
or  a  x  a  X  a  x  a,  or  aaaa  (Art.  12).     That  is,  if  a  =  4, 

4a  =  4  x  a  =  4  x  4  =  16, 

but    a4  =  a  x  a  X  a  X  a  =  4  x  4  x  4  x  4  =  256. 

18.  A  Term,  its  Dimensions,  and  Degree  —  Homo- 
geneous—  Similar.  —  A  term  is  an  Algebraic  expression 
in  which  no  two  of  the  parts  are  connected  by  the  sign  of 
addition  or  subtraction.  Thus  4a,  5a2bc,  and  Axy  -=-  bab  are 
terms.  2a,  4c2d,  and  —  bahl  are  the  terms  of  the  expression 
2a  +  4c2a"  —  5asd. 

Each  of  the  literal  factors  of  a  term  is  called  a  dimension 
of  the  term,  and  the  number  of  the  literal  factors  or 
dimensions  is  called  the  degree  of  the  term.  Thus  a?b3c 
w  aabbbc  is  said  to  be  of  six  dimensions  or  of  the  sixth 
degree,  because  it  contains  six  literal  factors,  viz.,  a  twice, 
b  three  times,  and  c  once.  A  numerical  coefficient  is  not 
counted  ;  thus  a2b3  and  ba2b8  are  of  the  same  degree,  i.e.,  the 
fifth  degree,  since  there  are  five  literal  factors,  viz.,  a  twice 
and  b  three  times. 

It  is  clear  that  the  degree  of  a  term,  or  the  number  of  its 
dimensions,  is  the  sum  of  the  exponents  of  its  literal  factors, 
provided  we  remember  that  if  no  exponent  be  expressed  1 
must  be  understood  (Art.  12).  Thus  a864ca  is  of  the  ninth 
degree,  since  3  +  4  +  2  =  9. 

Terms  are  homogeneous  when  they  are  of  the  same  degree. 
Thus  a3b,  <t~b'2,  :)abs,  are  homogeneous. 

Terms  are  similar  or  like  when  they  have  the  same  litem] 
part,  i.e.,  when  they  have  the  same  letters  and  the  corre- 
sponding Letters  affected  with  the  same  exponents.  Other- 
wise they  are  said  to  be  unlike.  Thus,  the  terms  8o'69, 
5a362,  and  —  u3b~  are  similar,  or  like;  but  the  terms  a2b  and 


SIMPLE  AND   COMPOUND   EXPRESSIONS.  13 

air  are  unlike,  since,  although  the  letters  are  the  same,  they 
are  not  raised  to  the  same  power. 

19.  Simple  and  Compound  Expressions.  —  A  simple 
expression  consists  of  only  one  term,  as  oab,  and  is  called  a 
monomial. 

A  compound  expression  consists  of  two  or  more  terms,  and 
is  called  a  polynomial,  or  multinomial. 

A  binomial  is  a  polynomial  of  two  terms.  Thus,  ab2  +  lac 
is  a  binomial. 

A  trinomial  is  a  polynomial  of  three  terms.  Thus,  a  -f-  b  —  c 
is  a  trinomial. 

A  polynomial  is  said  to  be  homogeneous  when  all  its  terms 
are  of  the  same  degree.  Thus  oab2  -f  7a?b  +  963  is  homo- 
geneous, for  each  term  is  of  the  third  degree. 

When  a  polynomial  consists  of  several  terms  of  different 
degrees,  the  degree  of  the  polynomial  is  that  of  its  highest 
term. 

A  polynomial  is  said  to  be  arranged  according  to  the 
powers  of  any  letter  it  contains  when  the  exponents  of  that 
letter  occur  in  the  order  of  their  magnitudes,  either  increasing 
or  decreasing.  Thus,  a4  +  4a86  +  6a2b'2  +  4a&3  is  arranged 
according  to  the  descending  powers  of  a,  and  4ab3  +  6a262 
+  4a36  +  a4  is  arranged  according  to  the  ascending  powers 
of  a. 

The  reciprocal  of  a  number  is  1  divided  by  that  number. 

Thus,  the  reciprocal  of  a  is  -.     If  the  product  of  two  num- 
a 

bers  is  1,  each  number  is  the  reciprocal  of  the  other. 

20.  Positive  and  Negative  Quantities.  —  In  Arith- 
metic we  deal  with  numbers  connected  by  the  signs  +  and 
— ,  and  in  finding  the  value  of  an  expression  such  as 
l_|_3_2  +  4we  understand  that  the  numbers  to  which 
the  sign  +  is  prefixed  are  additive,  and  those  to  which  the 
sign  —  is  prefixed  are  subtractive,  while  the  first  term,  1, 
to  which  no  sign  is  prefixed,  is  counted  among  the  additive 
terms.      The  same  thing  is  true  in  Algebra ;    thus  in  the 


14  POSITIVE  AND  NEGATIVE    QUANTITIES. 

expression  5a  +  lb  —  3c  —  2d  we  understand  the  s3'mbols 
5a  and  lb  to  be  additive,  while  3c  and  2d  are  subtractive. 

But  in  Arithmetic  the  sum  of  the  additive  terms  is  always 
greater  than  the  sum  of  the  subtractive  terms,  i.e.,  we  are 
always  required  to  subtract  a  smaller  number  from  a  greater  ; 
if  the  reverse  were  the  case  the  result  would  have  no  Arith- 
metic meaning,  i.e.,  we  could  not  in  Arithmetic  subtract  a 
greater  number  from  a  smaller.  In  Algebra,  however,  not 
only  may  the  sum  of  the  subtractive  terms  exceed  that  of 
the  additive,  but  a  subtractive  term  may  stand  alone,  and  yet 
have  a  meaning  quite  intelligible.  It  is  therefore  usual  to 
divide  all  Algebraic  quantities  into  x>ositive  quantities  and 
negative  quantities,  according  as  they  are  preceded  by  the 
sign  -f-  or  the  sign  —  ;  and  this  is  quite  irrespective  of  any 
actual  process  of  addition  and  subtraction. 

Illustration  (1)  Suppose  a  ship  were  to  start  from  the 
equator  and  sail  northward  100  miles  and  then  southward 
80  miles,  the  Algebraic  statement  would  be 

100  -  SO  =  +20. 

Here  the  positive  sign  of  the  result  indicates  that  the  ship  is 
20  miles  north  of  the  equator.  But  if  the  ship  first  sailed 
80  miles  northward  and  then  southward  100  miles,  the  Alge- 
braic statement  would  be 

80  -  100  =   -20. 

Here  the  negative  sign  of  the  result  indicates  that  the  ship 
is  20  miles  south  of  the  equator. 

(2)  Suppose  a  man  were  to  gain  $40  and  then  lose  $36, 
his  total  gain  would  be  $4.  But  if  he  first  gained  $36  and 
then  lost  $40,  he  sustained  a  loss  of  $4. 

The  corresponding  Algebraic  statements  would  be 

$40  -  $36  =  +$4, 
$36  -  $40  =  -$4. 

Here  the  negative  quantity  in  the  second  case  is  interpreted 
as  a  debt,  i.e.,  a  sum  of  money  opposite  in  character  to  the 


POSITIVE   AND  NEGATIVE    QUANTITIES.  15 

positive  quantity  or  gain  ill  the  first  case.  In  Arithmetic 
we  would  call  it  a  debt  or  loss  of  $4.  In  Algebra  we  make 
the  equivalent  statement  that  it  is  a  gain  of  —  $4. 

(3)  Suppose  a  man  starts  at  a  certain  point  and  walks 
100  yards  to  the  right  in  a  straight  line,  and  then  walks  back 
70  yards,  he  will  be  30  yards  to  the  right  of  his  starting 
point.  If  he  first  walks  from  the  same  point  70  yards  to 
the  right  and  then  walks  back  70  yards,  he  will  be  at  the 
point  from  which  he  started.  But  if  he  first  walks  to  the 
right  70  yards  and  then  walks  back  100  yards,  he  will  be 
30  yards  to  the  left  of  his  starting  point.  The  corresponding 
Algebraic  statements  are 

100  yards  —     70  yards  =       30  yards 
70      "      -    70     "      =        0      » 
70      «      -  100      "      =  -30      "    . 
Here  we  see  that  the  negative  sign  may  be  taken  as  indi- 
cating a  reversal  of  direction.     In  Arithmetic  we  would  say 
the  man  was  30  yards  to  the  left  of  his  starting  point.     In 
Algebra  we  say  he  was  —30  yards  to  the  right  of  his  start- 
ing point. 

There  are  numerous  instances  like  the  preceding  in  which  it  is  con- 
venient for  us  to  be  able  to  represent  not  only  the  magnitude  but  the 
nature  or  quality  of  the  things  about  which  we  are  reasoning.  As  in 
the  preceding  cases,  in  a  question  of  position  we  may  have  to  distin- 
guish a  distance  measured  to  the  north  of  the  equator  from  a  distance 
measured  to  the  south  of  it;  or  a  distance  measured  to  the  right  of  a 
certain  starting  point  from  a  distance  measured  to  the  left  of  it;  or 
we  may  have  to  distinguish  a  sum  of  money  gained  from  a  sum  of 
money  lost ;  and  so  on.  These  pairs  of  related  quantities  the  Alge- 
braist distinguishes  by  means  of  the  signs  +  and  — .  Thus  if  the 
things  to  be  distinguished  are  gain  and  loss,  he  may  denote  by  4  or  -f4 
a  gain,  and  then  he  will  denote  by  —4  a  loss  of  the  same  extent.  In 
this  way  we  can  conceive  the  possibility  of  the  independent  existence 
of  negative  quantities.  The  signs  -f  and  — ,  therefore,  are  used  to 
indicate  the  nature  of  quantities  as  positive  or  negative,  as  well  as 
to  indicate  addition  and  subtraction  (Arts.  8  and  9). 

In  Arithmetic  we  are  concerned  only  with  the  numbers 
which  begin  at  0  and  are  represented  by  the   symbols   0,  1, 


16  POSITIVE   AND   NEGATIVE   QUANTITIES. 

2,  3,  4,  etc.  without  limit,  and  intermediate  fractions.  But 
the  quantities  which  we  usually  measure  by  numbers  in  Alge* 
bra  do  not  really  begin  at  any  point,  but  extend  in  opposite 
directions  without  limit.  In  order  therefore  to  measure  such 
quantities  on  a  uniform  system,  the  symbols  of  Algebra  arc- 
considered  as  increasing  from  0  in  two  opposite  directions  ; 
i.e.,  besides  the  symbols  used  in  Arithmetic,  we  cousidei 
another  set  —1,  —  2,  —3,  —4,  etc.  without  limit,  and  inter- 
mediate fractions.  Symbols  iu  one  direction  are  preceded 
by  the  sign  +,  and  are  called  positive;  and  those  iu  the 
other  direction  are  preceded  by  the  sign  — ,  and  are  called 
negative.  Symbols  without  a  sign  prefixed  are  considered 
to  have  +  prefixed. 

These  two  sets  of  symbols  may  be  illustrated  as  follows  : 

...  -8,  -7,  -6,  -5,  -4,  -3,  -2,  -1,    0,    +1,  +2,  +3,  +4,  +5,  +6,  +7,  +8,  .  .  . 
l l l l l !_J l I I l l I l i L_J 

I 

the  positive  being  those  in  the  right  direction  from  zero, 
and  the  negative  those  in  the  left  direction  from  the  same 
point. 

Thus,  if  4  represent  a  distance  of  4  miles  measured  to  the  right  of 
a  certain  point,  —4  will  represent  a  distance  of  4  miles  measured  to 
the  left  of  the  same  point.  If  +4  represent  4  degrees  above  zero,  —4 
will  represent  4  degrees  below  zero.  If  +4  represent  4  years  after 
Christ,  —4  will  represent  4  years  before  Christ.  If  +4  represent  a 
fall  of  four  feet,  —4  will  represent  a  rise  of  4  feet.  If  +4  represent 
a  (jain  of  $4,  —4  will  represent  a  loss  of  $4.  In  general,  when  we 
have  to  consider  quantities  the  exact  reverse  of  each  other  in  their 
nature  or  quality,  we  may  regard  the  quantities  of  either  quality  as 
positive,  and  those  of  the  opposite  quality  as  negative.  It  matters 
not  which  quality  we  take  as  the  positive  one  so  long  as  we  take  the 
opposite  one  as  negative;  but  having  assumed  at  the  commencement 
of  an  investigation  a  certain  quality  as  positive,  the  important  point 
is  to  use  it  uniformly  and  consistently  throughout. 

The  absolute  value  of  any  quantity  is  the  number  repre- 
sented by  this  quantity  taken  independently  of  the  sign 
which  precedes  the  number.  Thus,  3  and  —3  have  the 
same  absolute  value. 


MULTIPLICATIONS   IN  ANY   ORDER.  17 

Negative  quantities  are  often  spoken  of  as  less  than  zero. 
For  example,  if  a  man's  debts  exceed  his  assets  by  $4,  it 
is  said  that  "he  is  worth  $4  less  than  nothing."  In  the 
language  of  Algebra  it  would  be  said  "  he  is  worth  —  $4." 

A  negative  number  is  said  to  be  Algebraically  greater  than 
another  when  it  is  numerically  less,  or  when  it  has  the  smaller 
absolute  value.  Thus  —  3  >  —  6,  since  —3  is  only  3  less 
than  0  while  —  G  is  6  less  than  0,  or  as  a  person  who  owes 
$3  is  better  off  than  one  who  owes  $6  ;  or  in  the  case  of  the 
thermometer,  when  the  mercury  is  at  10°  below  0  (marked 
—  10°)  at  one  hour,  and  at  —5°  at  another  hour,  the 
temperature  is  said  to  be  increasing;  i.e.,  —  5°  >  — 10°. 
Also,  in  Algebra,  zero  is  greater  than  any  negative  quantity, 
as  a  man  who  has  no  property  or  debt  is  considered  better 
off  than  one  who  is  in  debt.  Thus  it  is  easy  to  see  that 
in  the  series  on  page  16  each  number  is  greater  by  unity  than 
the  one  immediately  to  the  left  of  it. 

21.  Additions  and  Multiplications  may  be  Made 
in  any  Order. —  (1)  When  a  number  of  terms  are  con- 
nected by  the  signs  +  and  — ,  the  value  of  the  result  is  the 
same  in  whatever  order  the  terms  are  taken  ;  thus  6  +  5 
and  5  +  6  give  the  same  result  viz.,  11  ;  and  so  also  a  +  b 
and  b  +  a  give  the  same  result,  viz.,  the  sum  of  the  num- 
bers which  are  represented  by  a  and  b.  We  may  express 
this  fact  Algebraically  thus,  a  +  6  =  6  +  a.  Similarly 
a  —  b  +  c  =  a  +  c  —  b,  for  in  the  first  of  the  two  expres- 
sions b  is  taken  from  a,  and  c  added  to  the  result ;  in  the 
second  c  is  added  to  a,  and  b  taken  from  the  result. 

Similar  reasoning  applies  to  all  Algebraic  expressions. 
Hence  we  may  write  the  terms  of  an  expression  in  any  order 
we  please,  provided  each  has  its  proper  sign. 

Thus  it  appears  that  a  —  b  may  be  written  in  the  equiva- 
lent form  —  b  +  a.  As  an  illustration  we  may  suppose 
that  a  represents  a  gain  of  a  pounds,  and  —6a  loss  of  b 
pounds  ;  it  is  clearly  immaterial  whether  the  gain  precedes 
the  loss,  or  the  loss  precedes  the  sain. 


18         SUGGESTIONS  FOR   THE  STUDENT. 

(2)  When  one  number,  whether  integral  or  fractional,  is 
multiplied  by  a  second,  the  result  is  the  same  as  when  the 
second  is  multiplied  by  the  first. 

The  proof  for  whole  numbers  is  as  follows :  Write  down 
a  rows  of  units,  putting  b  units  in  each  row,  thus : 

I    |    |    |    | b  in  a  row, 

I    I    I    I    I  • 

I    I    I    I    I a  rows. 

Then  counting  by  rows  there  will  be  6  units  in  a  row  repeated 
a  times,  i.e.,  b  x  a  units.  Counting  by  columns  there  will 
be  a  units  in  a  column  repeated  b  times,  i.e.,  a  X  b  units. 

.-.  ba  =  ab. 

These  two  laws  are  together  called  the  Commutative  Laic, 
or  Law  of  Commutation. 

22.  Suggestions  for  the  Student  in  Solving  Ex- 
amples.—  In  solving  examples  the  student  should  clearly 
explain  how  each  step  follows  from  the  one  before  it ;  for 
this  purpose  short  verbal  explanations  are  often  necessary. 

The  sign  "  =  "  should  never  be  used  except  to  connect 
quantities  which  are  equal.  Beginners  should  be  particularly 
careful  not  to  employ  the  sign  of  equality  in  any  vague  and 
inexact  sense.  The  signs  of  equality,  in  the  several  steps 
of  the  work,  should  be  placed  one  under  the  other,  unless 
the  expressions  are  very  short. 

In  elementary  work  too  much  importance  cannot  be  at- 
tached to  neatness  of  style  and  arrangement.  The  beginner 
should  remember  that  neatness  is  in  itself  conducive  to 
accuracy. 

EXAMPLES. 

Find  the  numerical  value  of  the  following  expressions, 
when  a  =  1,  b  =  2,  e  =  3,  d  =  4,  and  e  =  5. 

1.  a2  +  b-  +  c9  +  d~  +  <-.  Ans.  55. 

2.  abc-  +  bed'1  —  dea".  94. 

3.  e*  -f  Mb2  +  64  -  4e"6  -  4e63.  81, 


EXAMPLES.  1^ 


4     ^       de_32^  Am^ 

4a        b2        o4 
.     8a2  +  362       4c2  +  662  _  tf_±&  15 


6. 


t2  +  &a  c2  -  b2  e 

a4  +  4a3&  +  6a262  +  ±ab3  +  b*  3 

a3  +  3a26  +  3ab2  +  &s 
7  V  +(*c  G. 

'      &2    +    d2    _    6d 

8.  (a  +  b)(b  +  c)-(b  +  c)(cA-d)  +  (c  +  d)(d  +  e).  43. 

9.  (a-2&  +  3c)2-(o-2c  +  3d)2-f(c-2a'  +  3e)2.    72. 

10.  V^c2  +  5d2  +  e).  11. 

11.  V^e2  +  «"2  +  c2  -  a2).  7. 
If  a  =  8,  b  =  6,  c  =  1,  a;  =  9,  y  —  4,  find  the  value  of 

14.  Find  the  difference  between  abx  and  a  +  6  +  a,-,  when 
a  =  5,  5  =  7,  and  oj  =  12.  Ans.  396. 

15.  When  a  =  3,  find  the  difference  between  a2  and  2a, 
a3  and  3a,  a4  and  4a,  a5  and  5a,  a6  and  Ga. 

Ans.  3,  18,  G9,  228,  and  711. 

16.  Find  the  value  of   3^  +  2as/(2a  +  b  -  x),  when 
a  =  6,  6  =  5,  c  =  4,  x  =  1.  ^l»s.  54. 

17.  Find  the  value  of  (9  -  y)(aj  +  l)  +  (®  +  5)  (y  +  7) 
—  112,  when  a;  =  3  and  y  =  5.  -4ns.  0. 

18.  Find  the  value  of  x\Ra?  -  8#)  +  y\R^  +  %)i  when 
a;  =  5  and  2/  =  3.  Ans.  20. 

19.  If  a  =  2,  6  =  3,  x  =  6,  and  ?/  =  5,  find  the  value  of 
y](a+byy\+f\(a  +  x)(y-2a)\+]/\(y-by-al       Ans.9. 

Find  the  value  of 

20.  a;4  -  11a;3  -  11a;2  -  13a;  +  11  for  x  =  12.  — 1. 

21.  x4  -  x3  —  4a;2  —  3x  —  5  for  x  =  3.  4. 

22.  a;5  -  3a;2  -  8  for  x  =  4.  968. 

23.  3a;4  -  60a;3  +  54a?  +  60a;  +  58  for  x  =  19.  115. 


20  EXAMPLES. 

24.  3x5'-  lG0.r4  +  344a3  +  700x2  -  1910a;  +  1200  for 
x  =  51.  Ans.  27. 

Express  the  following  in  Algebraic  symbols: 

25.  Seven  times  a,  plus  the  third  power  of  b.       la  +  b3. 

26.  Six  times  the  cube  of  a  multiplied  by  the  square  of  6, 
diminished  by  the  square  of  c  multiplied  by  the  fourth  power 
of  d.  Ans.  Ga3b2  -  cH\ 

27.  3  into  x  minus  m  times  y,  divided  by  m  minus  n. 

Ans.  (3x  —  my)  ~  (m  —  n) . 

28.  Four  times  the  fourth  power  of  a,  diminished  by  six 
times  the  cube  of  a  into  the  cube  of  b,  and  increased  by 
four  times  the  fourth  power  of  b.      Ans.  4a4  —  (Ja3b3  +  4b4. 

29.  Six  times  the  square  of  n,  divided  by  a  minus  b, 
increased  by  six  a  into  the  expression  m  plus  n  minus  b. 

Ans.  -^~  +  6a  (in  +  n  -  b). 
a  —  b 

30.  Three  times  the  square  root  of  x,  diminished  by  y 
mto  the  expression,  the  square  of  x  plus  the  product  of  a; 
and  y  plus  the  square  of  y,  and  increased  by  the  square  root 
of  x  into  the  expression  x  square  plus  y  square. 

Ans.  3\x  -  y(x-  +  xy  +  y-)  +  \fx(x-  +  y-). 


ADDITION  —  ALGEBRAIC   SUM.  21 


CHAPTER     II. 

ADDITION. 

23.  Addition  —  Algebraic  Sum.  —  Addition  in  Alge- 
bra is  the  process  of  finding  the  Algebraic  sum  of  several 
quantities.  The  Algebraic  sum  of  several  quantities  is 
their  aggregate  value,  and  it  is  usual  to  find  the  simplest 
equivalent  expression  for  it. 

It  is  convenient  to  make  three  cases  in  Addition  ;  (1)  when 
the  terms  to  be  added  are  like  (Art.  18),  and  have  like  signs; 
(2)  when  they  are  like  but  have  unlike  signs;  and  (3)  when 
they  are  unlike. 

24.  Case  1.  To  Add  Terms  which  are  Like  and 
have  Like  Signs.  —  Let  it  be  required  to  add  8x2y,  <±x'2y, 
and  lx2y. 

Here  8x2y  is  x2y  taken  8  times,  4x2y  is  x~y  taken  4  times, 
and  lx2y  is  x2y  taken  7  times  ;  therefore  x~y  is  taken  in  all 
8  +  4  +  7=19  times,  and  hence  the  sum  is  19an/. 

The  truth  of  this  will  be  evident  to  the  beginner  when  he 
remembers  that  the  three  quantities  8  lbs.,  4  lbs.,  and  7  lbs., 
udded  together,  give  19  lbs. 

Similarly  12ab  +  Sab  +  hab  +  ab  =  21ab. 

Let  it  be  required  to  add  —  Sab,  —lab.  and  —  9ab. 

Here  —  Sab  is  ab  taken  —3  times,  —  lab  is  ab  taken  —7 
times,  and  —  9ab  is  ab  taken  —9  times  ;  therefore  ab  is  taken 
in  all  —19  times,  and  hence  the  sum  is  — 19«&. 

The  truth  of  this  will  be  evident  from  the  consideration  that, 
if  a  sum  of  money  be  diminished  successively  by  $3,  $7,  and 
$9,  it  is  diminished  altogether  by  $19. 

Therefore,  to  add  like  terms  which  have  the  same  sign,  add 
the  numerical  coefficients,  prefix  the  common  sign,  and  annex 
the  common  symbols. 


22  TO  ADD   LIKE   TERMS    WITH   UNLIKE  SIGNS. 

For  example,  Ga  +  3a  -f  a  +  la  =  17a,  and  —  2a6  —  lab 
-9ab  =  -18a&. 

25.  Case  2.  To  Add  Terms  which  are  Like,  but 
have  Unlike  Signs.  —  Let  it  be  required  to  add  9a  and 
-4a. 

Here  —4a  destroys  4  of  the  9  times  a,  and  gives  when 
added  to  it,  5a.  This  is  usually  expressed  by  saying  —4a 
will  cancel  -f  4a  in  the  term  da,  and  leave  +5a  for  the  aggre- 
gate or  sum  of  the  two  terms.  # 

For  if  9a  denote  $9  which  a  man  has  in  his  possession, 
and  —4a  denote  a  debt  of  $4,  then  the  aggregate  value  of 
his  money  is  $5. 

In  like  manner  if  it  be  required  to  add  8a,  —9a,  —a,  3a, 
4a,  —11a,  a,  we  find  the  sum  of  the  positive  terms  to  be 
16a,  and  the  sum  of  the  negative  terms  to  be  —21a;  now 
+  16a  will  cancel  —16a  in  the  term  —21a,  which  leaves  —5a 
for  the  aggregate  or  sum  of  the  terms. 

Therefore,  to  add  like  terms  which  have  not  all  the  same 
sign,  add  all  the  positive  numerical  coefficients  into  one  sum, 
and  all  the  negative  numerical  coefficients  into  another ;  take 
the  difference  of  these  tivo  sums,  prefix  the  sign  of  the  greater, 
and  annex  the  common  symbols. 

For  example  7a  —  3a+lla-f a  —  5a  —  2a  — 19a  —  10a  =  9a, 
and  5a6-6a6+2ao  —  lab  — 3a&+4a&=llao-16a&  =  -5a&. 

We  need  not,  however,  strictly  adhere  to  this  rule,  for 
since  terms  may  be  added  or  subtracted  in  any  order  (Art. 
21),  we  may  choose  the  order  we  find  most  convenient. 

Thus,  in  the  last  example,  we  may  say  5a&  added  to  —Gab 
gives  —  ab ;  adding  —  ab  to  +2a&  gives  +a&;  adding  +ab 
to  —  lab  gives  —Sab;  adding  —  Gab  to  —3ab  gives  —  9a& ; 
adding  —9db  to  +  \ah  gives  —  5a&,  for  the  sum,  which  is  the 
same  as  was  found  by  the  rule.' 

26.  Case  3.  To  Add  Terms  which  are  not  all 
Like  Terms.  —  Let  it  be  required  to  add  la  +  bb  —  1c 
+  :\d,  3a  —  b  +  2o  +  5a*,  9a  —  2b  —  c  —  d,  and  —a  +  36 
+  4c  -  3d  +  e. 


TO  ADD  TERMS  WHICH  ARE  NOT  ALL  LIKE  TERMS.      23 

It  is  convenient  to  arrange  the  terms  in  columns,  so  that 
like  terms  shall  stand  in  the  same  column  ;  and  then  add 
each  column,  beginning  with  that  on  the  left,  as  follows  : 

4a  +56  -7c  +3d* 
3a  —  6  +  2c  +  5d 
9a  —  2b  —  c  —  d 
~a  +36  +4c  -3a"  +e 

15a  +  56  -2c  +4d  +e 

Here  the  terms  4a,  3a,  9a,  and  —a  are  all  like  terms ;  the 
sum  of  the  positive  terms  is  16a ;  there  is  one  negative  term, 
viz.,  —a,  so  that  the  sum  of  the  terms  in  the  first  column  by 
Art.  25  is  +15a;  the  sign  +  may  be  omitted  by  Art.  9. 
Similarly  5b  —  b  —  2b  +  3b  —  56,  —  7c  +  2c  -  c  +  4c  =  -  2c, 
and  so  on  ;  there  being  no  term  similar  to  e,  it  is  connected 
to  the  other  terms  by  its  proper  sign. 

Therefore,  to  add  terms  which  are  not  all  like  terms,  add 
together  the  terms  which  are  like  terms,  by  the  ride  in  Case  2, 
and  set  doicn  the  other  terms  each  preceded  by  its  proper  sign. 
In  the  two  following  examples   the  terms  are  arranged 
suitably  in  columns. 

x3  +2x2  -  3a;  +1  a2  +  db  +  lr  -c 

4a;3  +7x2  +     a;  -9  3a2  — 3a6  —7b2 

-2x3  +  x2  -   9a;  +8  4a2  +5a6  +962 

-3x8  -  x2  +10aj  -1  a2  -3a6  -362 


9a;2  _     x  __i  9ft2  _c 

In  the  first  example  we  have  in  the  first  column  x3  +  4teB 
—  2a;3  —  3a;3  =  5a;3  —  5a;3  =  0 ;  this  is  usually  expressed  by 
saying  the  terms  which  involve  xs  cancel  each  other. 

Similarly,  in  the  second  example,  the  terms  which  involve 
ab  cancel  each  other ;  and  also  those  which  involve  6a  cancel 
each  other. 

27.  Remarks  on  Addition.  —  "We  have  seen  that  when 
two  or  more  like  terms  are  to  be  added  together  they  may 
be  collected  into  a  single  term  (Arts.  24  and  25).     If,  how-. 


24  REMARKS    ON  ADDITION  —  EXAMPLES. 

ever,  the  terras  are  unlike  they  cannot  be  collected.  Thus 
we  write  the  sum  of  a  and  b  in  the  form  a  +  b,  and  the  sum 
of  a  and  —  b  in  the  form  a  —  b. 

From  the  foregoing  examples  it  will  be  observed  that  in 
Algebra  the  word  sum  is  used  in  a  wider  sense  than  in 
Arithmetic.  Thus,  in  the  language  of  Arithmetic,  a  —  b 
signifies  that  b  is  to  be  subtracted  from  a,  and  has  no  other 
meaning  ;  but  in  Algebra  it  also  means  the  sum  of  the  two 
quantities  a  and  —b  without  any  regard  to  the  relative 
magnitudes  of  a  and  6. 

When  quantities  are  connected  by  the  signs  -f-  and  — ,  the 
resulting  expression  is  called  their  Algebraic  sum.  Thus 
the  Algebraic  sum  of  12a  —  29a  +  14a  is  —3a. 

In  Algebra,  wherever  the  word  sum  is  used  without  an 
adjective,  the  Algebraic  sum  is  understood. 

EXAMPLES. 


1. 

2. 

3. 

4. 

4aa; 

Gab 

26a;2- 

2a?b 

bax 

-lab 

Sbx2 

-   an-b 

—  3  ax 

-Sab 

—  9&a;2 

11  a2b 

2ax 

bab 

Ibx2 

-ba*b 

—  7  ax 

-dab 

2bx2 

4a2b 

ax 

2ab 

—Abx- 

-da-b 

2ax 

-6ab 

-56»9 

2a2b 

b. 

7a;2 

-3xy 

+     x 

3a;2 

-  f 

+  3a:  - 

-  V 

-2a2 

+  4.T?/   +  b>f 

—     x  - 

-ty 

—  "•<'.'/ 

-  f 

+  9x  - 

-  :>// 

4a2 

+4/ 

-   2x 

12a;2  —  6xy  +ly2  +  10a;  —8y 
Add  together  the  following  expressions  : 

6.  a+2b  —  3c,  —3a  +  b-\-2c,  2a— 36+c.  Ans.  0. 

7.  3a+26  —  c,  -a+36+2c,  2a-6+8c.         4o+46+4c 


EXAMPLES. 


25 


8.  -3x+2y+z,  x-3y  +  2z,  2x+y—3z.  Ans.  0. 

9.  -X+2y+3z,  Bx-y+2z,  2x+3y—z.  4x+±y+4z. 
10.   4a+36+5c,  —  2a+36— 8c,  a  —  6+c.          3a+56— 2c. 

$s  11.    — 15a-196-18c,  14a  +  14&+8c,  a  +  hb  +9c.  -c. 

12.  25a  — 156+c,  13a  — 106+4c,  a+206-c.  39a— 56+4c. 

13.  _16a-106+5c,  10a+5&+c,  6a+5&-c.  5c. 
In  adding  together  several  expressions  containing   terms 

with  different  powers  of  the  same  letter,  it  will  be  found 
convenient  to  arrange  all  the  expressions  in  ascending  or 
descending  powers  of  that  letter  (Art.  19). 
U.    3x3+7-5x2,  2a,-2- 8 -9a,-,  4x-2x3+3x\ 
Arranging  the   terms  in  the  descending  powers  of  x,  we 

have 

3a;3  -5a;2  +7 

2ar  -9a;  -8 
—  2a;3  +3a/2  +4a; 

a;8  —ox  —1  Ans. 

15.  3a&2-268+as,    5a2&-a£/2-3a3,    8a3+5&3,    9a2Z>-2a3 
+ab\  Ans.  363+3a&2+14a2&  +  4«3. 

It  will  be  observed  that  this  answer  is  arranged  according 
to  descending  powers  of  &,  and  ascending  powers  of  a. 

16.  2x*-2xy+3y*,  ±y2+oxy-2x2,  x2-2xy-Gy2. 

Ajis.  x2+xy+y2. 

17.  a8— a2+3a,  3a3+4a2+8a,  5a8— 6a2— 11a. 

Ans.  9a3 -3a2. 

18.  x*+Sx*y+3xy2,  -3x2y-6xy2-x3,  3x2y+4:Xy2. 

Ans.  3x2y+xy2. 

19.  a;8— 2aa2+a2a;+a8,  a;3+3aar,  2a3 -ax2 -2x3. 

Ans.  a2a?-r-3a8. 

20.  2a&  -  Sax2  +  2a2a;,    1 2a&  +  1  Oaa;2  -  Ga2.«,   -  8a&  +  ax3 
—  bct-x.  Ans.  Gab  —  9a2x+7ax2+ax3. 

21.  x2+*f+z3,  -4a;2-523,  8ar-7?/4+10z3,  6y4-6z3. 

Ans.  ox2. 

22.  a;4-4a;3?/-f6xV-4.^3+?y4,    Ax3y- l2xY+12xy3-4y\ 
6x2y2-12xy3+Gy\  -ixy3—iy\  y\  Ans.  xK 


26  SUBTRACTION  —  ALGEBRAIC  DIFFERENCE. 


CHAPTER     III. 

SUBTRACTION. 

28.  Subtraction  —  Algebraic  Difference.  —  Subtrac- 
tion in  Algebra  is  the  process  of  finding  the  difference 
between  two  Algebraic  quantities. 

The  Algebraic  Difference  of  two  quantities  is  the  number 
of  units  which  must  be  added  to  one  in  order  to  produce  the 
other.  Thus,  what  is  the  difference  between  2  and  G  means 
"  how  many  units  added  to  2  will  make  G  "  ?  The  Difference 
is  sometimes  called  the  Remainder. 

The  Subtrahend  is  the  quantity  to  be  subtracted  ;  or  it  is 
the  one  from  which  we  measure.  Thus,  2  is  the  subtrahend 
in  the  above  example. 

The  Minuend  is  the  quantity  from  which  the  subtrahend  is 
taken  ;  or  it  is  the  one  to  which  we  measure.  Thus,  G  is  the 
minuend  in  the  above  example. 

If  the  minuend  is  Algebraically  greater  than  the  subtra- 
hend, the  difference  is  positive  (Art.  20). 

If  the  minuend  is  Algebraically  less  than  the  subtrahend, 
the  difference  is  negative. 

In  Arithmetic  we  cannot  subtract  a  greater  number  from 
a  less  one,  because  subtraction  in  Arithmetic  means  taking  a 
less  number  from  a  greater.  But  in  Algebra  there  is  no  such 
restriction,  because  Algebraic  subtraction  means  finding  a 
dfference. 

29.  Rule  for  Algebraic  Subtraction.  —  Let  distances 
to  the  right  of  the  zero  point  be  called  positive,  and  those  to 
the  left  of  the  same  point  be  called  negative  (Figure,  Art. 
20).  Also  call  measuring  toward  the  right  from  any  point 
positive,  and  measuring  toward  the  left  from  any  point 
negative. 


RULE  FOR  ALGEBRAIC  SUBTRACTION.  27 

Then  the  difference  between  2  and  6  means  either  how 
many  units  must  toe  measure,  and  in  what  direction,  in  order 
to  pass  from  2  to  6  or  to  pass  from  6  to  2.  In  the  first 
case  we  begin  at  2  and  measure  four  units  to  the  right  and 
say  2  from  6  is  +4.  In  the  second  case  we  begin  at  G 
and  measure  four  units  to  the  left  and  say  6  from  2  is  —4. 
That  is,  if  we  subtract  2  from  6  the  difference  is  4  ;  but  if 
we  subtract  6  from  2  the  difference  is  —4. 

Also  to  find  the  difference  between  —1  and  +1,  we  may 
begin  at  —1  and  measure  2  units  to  the  right  and  get  +2, 
or  we  may  begin  at  +1  and  measure  2  units  to  the  left  and 
get  —2  ;  i.e.,  if  we  subtract  —  1  from  +1  the  difference  is  +2, 
but  if  we  subtract  +1  from  —1  the  difference  is  —2. 

Similarly  the  difference  between  —2  and  —  7  is  —  5  or 
+  5,  according  as  we  measure  from  —2  toward  the  left 
to  —7  or  from  —7  toward  the  right  to  —2;  i.e.,  if  we 
subtract  —2  from  —7  the  remainder  is  —5,  but  if  we  sub- 
tract —  7  from  —2  the  remainder  is  +5.  And  also,  the 
difference  between  —6  and  +7  is  +13  or  —13  according 
as  we  measure  from  —6  to  +7  or  from  +7  to  —  6  ;  i.e.,  if 
we  subtract  —  6  from  +7  the  difference  is  13,  but  if  we 
subtract  7  from  —  6  the  difference  is  —13. 

Hence  we  see  that  the  remainder  in  each  case  is  found  by 
changing  the  Algebraic  sign  of  the  subtrahend,  and  then 
adding  it  Algebraically  to  the  minuend. 

Otherwise  thus.  Suppose  we  have  to  take  9  +  5  from  16  ; 
the  result  is  the  same  as  if  we  first  take  9  from  16,  and  then 
take  5  from  the  remainder ;  that  is,  the  result  is  denoted  by 
16-9-5. 

Thus  16  -  (9  +  5)  =  16  -  9  -  5. 

Here  we  enclose  9  +  5  in  parenthesis  in  the  first  expres- 
sion, because  we  are  to  take  the  whole  of  9  +  5  from  16 
(Art.  15). 

Suppose  we  have  to  take  9  —  5  from  16.  If  we  take  9 
from  16,  we  obtain  16  —  9  ;  but  we  have  thus  taken  too 
much  from  16,  for  we  had  to  take,  not  9,  but  9  diminished 


28  RULE   FOR  ALGEBRAIC  SUBTRACTION. 

by  5.  Hence  we  must  increase  the  result  by  5  ;  and  thus  we 
obtain  16  -  ((J  -  5)  =  1G  -  9  +  5. 

Similarly,  1G  —  (G  +  4  -  1)  =  16  -  6  -  4  +  1. 

In  like  manner  suppose  we  have  to  subtract  b  —  c  from  a. 
If  we  subtract  b  from  a,  we  obtain  a  —  b  ;  but  we  have  thus 
taken  too  much  from  a,  for  we  are  required  to  take,  not  b, 
but  b  diminished  by  c.  Hence  we  must  increase  the  result 
a  —  b  by  c ;  and  thus  we  obtain  a  —  (b  —  c)  =  a  —  b  +  c 
for  the  true  remainder. 

Similarly,  a  —  (b  +  c  —  d)  =  a  —  b  —  c  +  d. 

Suppose  we  have  to  subtract  b  —  c  +  d  —  e  from  a.  This 
is  the  same  thing  as  subtracting  b  +  d  —  c  —  e  from  a  (Art. 
21).  If  we  subtract  b  -\-  d  from  a,  we  obtain  a  —  b  —  d; 
but  we  have  thus  taken  too  much  from  o,  for  we  were  to 
take,  not  b  +  d,  but  b  +  d  diminished  by  c  and  e.  Hence 
we  must  increase  the  result  by  c  +  e,  and  thus  obtain 

(6—  (b  —  c  +  cl  —  e)  =  a  —  b  —  d+c+e  =  a—  b+c  —  d+e. 

From  considering  each  of  these  examples,  it  is  evident 
that  subtracting  a  positive  number  is  the  same  thing  as  adding 
an  equal  negative  number,  and  also  that  subtracting  a  negative 
number  is  the  same  thing  as  adding  an  equal  positive  number. 
Therefore,  Algebraic  subtraction  is  equivalent  to  the  Alge- 
braic addition  of  a  number  with  the  opposite  Algebraic  sign. 

Hence  for  subtraction  we  have  the  following 

Rule. 
Change  the  signs  of  all  the  terms  in  the  subtrahend,  and 
then  add  the  result  to  the  minuend. 

EXAMPLES. 

1 .   Let  it  be  required  to  subtract  3a;— y+z  from  4aj— Sy+2z. 

Changing  the  signs  of  all  the  terms  in  the  sulitrahend.it 
stands  as  follows:  —  3a;  +  y  —  Z.  Then  collecting  as  in 
addition,  we  have 

4a;  —  3y  +  2z  —  ox  +  y  —  z  =  x  —  2y  -f  z. 


EXAMPLES.  29 

2.  From  3a;4  +  5a;3  —  6sca  —  Ix  +  5 

take  2a;4  —  2a;3  +  5ar  —  6aj  —  7. 
Changing  the  signs  of  all  the  terms  in  the  subtrahend,  and 
proceeding  as  in  addition,  we  have 

3a;4  +  5a;3  -     6a;2  -  7x  +     5 
-2x*  +  2a;3  -     5a;2  +  6x  +     7 


a;4  +  7a-3  -  lis2  -     x  +  12 

Kem.  —  The  beginner  may  solve  a  few  examples  by  actually  changing 
the  signs  of  the  subtrahend  and  going  through  the  operation  as  fully  as 
we  have  done  in  these  two  examples;  but  he  may  gradually  accustom 
himself  to  perform  the  subtraction  without  actually  changing  the  signs, 
but  merely  changing  them  mentally,  as  in  the  following  example. 

3.  From  Sab  +  lac  +  2c2  take  oab  —  4ac  +  3c2  —  d. 

Writing  the  subtrahend  under  the  minuend  so  that  similar 

terms  shall  fall  in  the  same  column,  for  convenience  (Art. 

26),  we  have  0  7    .      „       .    0  .-, 

''  Sab  -f     <ac  +  2c 

5ab  —    \.ac  +  3c2  —  d 


oab  +  llac  —    c2  +  cl 


Changing  the  sign  of  5ab  from  +  to  —  and  adding  it  to 
8ab,  we  have  Sab  ;  in  like  manner,  changing  the  sign  of  —  4ac 
from  —  to  +  and  adding  it  to  lac,  we  have  llac;  also 
changing  the  sign  of  +3c2  from  +  to  —  and  adding  it  to 
2c2,  we  have  —  c2 ;  changing  the  sign  of  —  d  and  adding  it, 
we  have  +d. 

Every  example  in  subtraction  may  be  verified  by  adding 
the  remainder  to  the  subtrahend ;  the  sum  will  be  equal  to 
the  minuend. 

30.  Remarks  on  Addition  and  Subtraction.  —  In 
Arithmetic  addition  always  produces  increase  and  subtrac- 
tion decrease;  but  in  Algebra  addition  may  produce  decrease 
and  subtraction  may  produce  increase.  Thus  in  Algebra  we 
may  add  —4a  to  8a  and  obtain  the  Algebraic  sum  4a,  which 
is  smaller  than  8a;  or  we  may  subtract  —3a  from  oa  and 
obtain  the  Algebraic  difference  8a,  which  is  larger  than  5a. 


30  EXAMPLES. 


EXAMPLES. 

1.    From  5a;2  +    xy  —  3if 

Subtract  2a;2  +  8xy  -  ly1 


Remainder 

3a:2  —  Ixy  +  Ay'2 

2.    From 

x*  -  2a;8             -    9aj  +    4 

Subtract 

2a;4              -  3a;2  +     7x  -    8 

Remainder         -  a;4  -  2a;3  +  3a;2  -  16a;  +  12 
From 

3.  15as  +  lOy  -  18z  subtract  2x  -  8y  +  z. 

Ans.  13a;  +  18y  -  19s. 

4.  x  —  y  —  z  subtract  —10a;  —  14?/  +  15z. 

Ans.  11a;  +  13y  —  16z. 

5.  25a  —  166  —  18c  take  4a  -  36  +  15c. 

Ans.  21a  -  136  -  33c. 

6.  yz  —  zx  -(-  xy  take  —  xy  +  yz  —  zx.  2xy. 

7.  -2a;3  -  a;2  -  3a;  +  2  take  x3  -  x  +  1. 

Jlns.  —3a;3  —  a;2  —  2a;  +  1. 

8.  4a;2  -  3a;  +  2  take  -5a;2  +  6a;  -  7.      9a;2  -  9a;  +  9. 

9.  a;3  +  Ha3  +  4  take  8a;2  -  5a;  -  3.     a-3  +  3ar  +  5a;  +  7. 
10.    -8aV2  +  5a;2  +  15  take  9a2ar  -  8a;2  -  5. 


Ans.  -17a2a;2  +  13a;2  4-  20 

11. 

^2- 

-  $®y  -  \f  take  -fa;2  +  xy  -  if. 

Ans.  2x2  —  $«y  —%y° 

12. 

frf  - 

-  f a  -  1  take  -fa2  ■+■  a  —  $.     fa3  -  fa  -  £ 

13. 

&  - 

-  $#  +  |  take  |a;  -  1  +  \x\     -\a?  -  fa  +  £ 

14. 

|rf  - 

-  %ax  take  $  —  £a;2  —  faa;.          fa;2  +  ^o.r  —  £ 

15. 

fa;8- 

-  &y*  -  f  take  \xhj  -  £  if  -  \xy\ 

Ans.  fa;8  -  ^xhj  -  £?/2. 

31.  The  Use  of  Parentheses.*— A  parenthesis  indicates 
that  the  terms  enclosed  within  it  are  to  be  considered  as  one 
quantity  (Art.  15).     On  account  of  the  extensive  use  which 


*  As  the  bracket,  brace,  bar,  and  vinculum  all  have  the  Bame  significance  as  the 
parenthesis  (Art.  15),  the  rules  for  their  removal  or  introduction  are  the  same. 


MINUS   SIGN   BEFORE    THE   PARENTHESIS.  31 

is  made  of  parentheses  in  Algebra,  it  is  necessary  that  the 
student  should  become  acquainted  with  the  rules  for  their 
removal  or  introduction. 

32.  Plus  Sign  before  the  Parenthesis.  —  When  a 
2~>arenthesis  is  preceded  by  the  sign  -f ,  the  parenthesis  can  be 
removed  ivithout  making  any  change  in  the  expression  within 
the  parenthesis. 

This  rule  has  already  been  illustrated  in  Arts.  25  and  26  ; 
it  is  in  fact  the  rule  for  addition. 

7  +  (12  +  4)  means  that  12  and  4  are  to  be  added  and 
their  sum  added  to  7.  It  is  clear  that  12  and  4  may  be 
added  separately  or  together  without  altering  the  result. 

Thus         7  +  (12  +  4)  =  7  +  12  +  4  =  23. 

Also  a  +  (b  +  c)  means  that  b  and  c  are  to  be  added 
together  and  their  sum  added  to  a. 

Thus  a  +  (b  +  c)  =  a  +  b  +  c. 

7  +  (12  —  4)  means  that  to  7  we  are  to  add  the  excess 
of  12  over  4 ;  now  if  we  add  12  to  7,  we  have  added  4  too 
much,  and  must  therefore  take  4  from  the  result. 

Thus         7  +  (12  -  4)  =  7  +  12  -  4  =  15. 

Similarly  a  +  (b  —  c)  means  that  to  a  we  are  to  add  b 
diminished  by  c. 

Thus  a  +  (b  —  c)  =  a  +  b  —  c. 

Therefore  Conversely  :  Any  part  of  an  expression  may  be 
enclosed  within  a  parenthesis  and  the  sign  -f-  placed  before  it, 
the  sign  of  every  term  within  the  parenthesis  remaining  un- 
altered. 

Thus,  the  expression  a  —  b  +  c  —  rf  +  e  may  be  written 
in  any  of  the  following  ways  : 

a-6+c+(-d+e),  a-6+(c-d+e),  a+(— b+c— d+e), 

and  so  on. 

33.  Minus  Sign  before  the  Parenthesis.—  When  a 
parenthesis  is  preceded  by  the  sign  — ,  the  parenthesis  may 
be  removed  if  the  sign  of  every  term  within  the  parenthesis  be 
changed. 


32  COMPOUND    PARENTHESES. 

This  rule  has  already  been  illustrated  in  Art.  29  ;  it  is  in 
fact  the  rule  for  subtraction.  The  rule  is  evident,  because 
the  sign  —  before  a  parenthesis  shows  that  the  whole  ex- 
pression within  the  parenthesis  is  to  be  subtracted,  and  the 
subtraction  is  effected  by  changing  the  signs  of  all  the  terms 
of  the  expression  to  be  subtracted. 

Thus  a  —  (b  +  c)  =  a  —  b  —  c. 

Also  a  —  (b  —  c)  =  a  —  b  +  c. 

Therefore  Conversely  :  Any  part  of  an  expression  may  be 
enclosed  within  a  p>arenthesis  and  the  sign  —  placed  before  it, 
provided  the  sign  of  every  term  within  the  p>arenthesis  be 
changed.  The  proof  of  this  operation  is  to  clear  the  paren- 
thesis introduced,  and  thus  obtain  the  original  expression. 

Thus  a  —  b  +  c  +  d  —  e  may  be  written  in  the  following 
ways: 
a-b+c—  (-cZ+e),  a—b—  (— c  — d+e),  a—(b-c— d+e), 

and  so  on. 

34.  Compound  Parentheses.  —  Expressions  may  occur 
with  more  than  one  pair  of  parentheses ;  these  parentheses 
may  be  removed  in  succession  by  the  preceding  rules. 

We  may  either  begin  with  the  outside  parenthesis  and  go 
inward,  or  begin  with  the  inside  parenthesis  and  go  outward. 
It  is  usually  best  to  begin  with  the  inside  parenthesis.  The 
beginner  is  recommended  always  to  remove  first  the  inside 
pair,  next  the  inside  of  all  that  remain,  and  so  on.  Thus 
for  example  ; 

a  +  \b  +  (c  -  d)  \  =  a  +  ]b  +  c  -  d\  =  a  +  b  +  c  -  d. 
a  +  \b  —  (c  —  d)  \  =  a  +  $6  —  c  +  d\  =  a  +  6  —  c  +  d. 
a  -  \b  +  (c  -  d)  \  =  a  -  \b  +  c  -  d\  =  a  -  b  -  c  +  d. 
a  -  \b  -  (c  -  d)  \  =  a  -  \b  -  c  +  d\  =  a  -  b  +  c  -  d. 

It  will  be  seen  in  these  examples  that,  to  prevent  confusion 
between  different  pairs  of  parentheses,  we  employ  those  of 
different  forms;  and  hence  we  use.  besides  the  parenthesis, 
the  brace,  the  bracket,  and  sometimes  the  vinculum  (Art.  1 5  ) . 


COMPOUND 

PARENTHESES  — EXAMPLES. 

33 

Thus,  for  example, 
a  —  [6  —  \  c  —  (d  —  e~ 
=  a  — [6  — {c  — (Z-f 

=  a  —  &  +  c  —  d  + 

■«-/■ 

c  +  d  - 

-  e  +  /] 

Also 
-26 

-  [4a  -  66  - 

^3a  — c  +  (5a  —  26  - 

a 

-  3a  —  c 

:  +  2&)fj 

=  a  -  26  -  [4a  -  66  -  { 3a  -  c  +  (5a  -  26  -  3a  +  c  -  26)  j  ] 
=  a  _  26  -  [4a  -  66  -  \3a  -  c  +  5a  -  26  -  3a  +  c  -  26 }] 
_  a  _  26  -  [4a  -  66  -  3a  +  c  -  5a  +  26  +  3a  -  c  +  26] 
_  a  -  26  -  4a  +  66  +  3a  —  o  +  5a  —  26  —  3a  +  c  —  26 
=  2a,  by  collecting  like  terms. 

EXAMPLES. 

Simplify  the  following  expressions  by  removing  the  paren- 
theses and  collecting  like  terms. 

1.  a  — (6  — c)  +  a  +  (6  — c)+6— (a+c).    Ans.  a+b—c. 

2.  a  -  [6  +  \a  -  (6  +  a)\].  a. 
J)  a  -  [2a  -  ^36  -  (4c  -  2a) \].  a  +  36  -  4c. 
(1t>   \a.-(b-c)\  +  \b-(c-a)\-\c-(a-b)\.   3a-6-c. 

5.  2a-(56  +  [3c-a])-(5a-[6+c]).     -2a-46-2c. 

6.  -\a-\b-(c-a)X]-\b-\c-(a-b)X\.     b-a. 

~D  -(-(-(-*)))-(-(- *))■  «-  y. 

8.  _[5a5  _  (iij,  _  3a;)]  -  [5y  -  (3a  -  6?/)].      -5a;. 

9.  -[15a  -  \Uy  -  (15z  +  12?/)  -  (10a;  -  15s) J]- 

J.ws.  —25a;  +  2y. 

10.  8a;—  |16y  —  [3a;-(12j/  —  x)  —  8y']  +  x\.     11a;- 36?/. 

11.  -  [x  -  \z  +  (a;  -  s)  —  (s  —  a;)  —  sj  -  a?].    2x  -  2s. 

12.  —[a  4-  \a  -  («  —  »)  —  («  +  »)  -  aj  —  «]•       2«. 

13.  -[a  —  ja  +  (a  -  a)  —  («  —  a)  —  a|  -  2a].      a. 

14.  2a  -  [2a  -  ^2a  -  (2a  -  2a  —  a)}].  a. 

15.  16  -  x  —[7x  -  ^8a;  -  (9jb  -  3a;  —  6a;)  |].   16  -  12a;. 

16.  2a;  -  [3?/  -  |4a>  -  (by  -  6a;  -  7t/)|].     12a;  -  15?/. 

17.  2a-[36  +  (25-c)-4c  +  j2a-(3&-c=26)|].    4c. 

18.  a-  [56-  {a-  (5c-2c^6-46)  +  2a  -  (a  -  26  +  c)  \]. 

Ans.  3a  —  2c. 


34  EXAMPLES. 

19.  jb»_  [4.1;3_  s,  6aJ2_  (4iC  _  ! )  |]  _  (.1.4  +  4r3_}_fil.2+47.  +  j  ^ 

.4?is.  —  <Sa-3  —  8sb. 

When  the  beginner  has  had  a  little  practice  the  number 

of  steps  may  be  considerably  diminished ;    he  may  begin  at 

the  outside  and  remove  two  or  more  parentheses  at  once,  as 

follows  : 

20.  a  -  [26  +  \Sc  -3a-(<i  +b)\  +  2a  -  (6  +  3c)] 
=    a  —  26  —  3c  +  3a  +  a  +  b  -  2a  +  b  +  3c 

=  3a. 

21.  a-{b-c)  -[a-b-c-2\b  +  c-S(c-a)-d\l. 

Ans.  6a  +  26  -  2c  -  2d. 

22.  2x-  (3y  -  4z)  -  \2x  -  (3?/  +  4z)  \  -  \dy  -  (4a  +  2a?) }. 

,4ns.  2x  —  3y  +  12z. 

23.  -20  (a  -  a")  +  3(6  -  c)  -  2[6  +  c  +  d  -  3|c  +  d 
_  4(a-  _  a)|].  ,4ns.  4a  +  b  +  c. 

24.  -4(a  +  d)  +  24(6  -  c)  -  2[c  +  d  +  a  -  3f<2  +  a 
-  4(6  +  c)H-  ^s-  -50c- 

25.  2(36  -  5a)  -  7[a  -  6^2  -  5 (a  -  6)H- 

.4ns.  -227a  +  2166  +  84. 

26.  -10ja-6[a-(&-c)]J  +  60f&-(c  +  a)j.     -10a. 

27.  _3^-2[-4(-a)]^  +  5^-2[-2(-a)]f.  4a. 

28.  _2|-[-(«  -  y)]J  +  $-*[-(«  -  *)]*.  0. 

Note.  —  The  line  between  the  numerator  and  denominator  of  a 
fraction  is  a  kind  of  vinculum.    Thus  is  equivalent  to  J(o:  —  3). 

»•  ii-.(»-)i-i{K»-f)-f[«-K-T)]r 

,4ns.  -Va  —  26. 

r«v_^«        i   i  *   .  j_ 

30. 

5  10 (  7' 

ylns.  12.r  —  3Qy. 


35[^-i{8«-|(7»-4,)}]+8(y-2ai). 


«■  S{|o -»>-•<»- •>}-{*-? 


-(a  -6)1.  Ans.  a  -  y-6  +  V1 


EXAMPLES.  35 

»■  HMHHHH')}-^)) 

./Ins.  0. 
The  terms  of  an  expression  can  be  placed  in  parentheses 
in  various  ways  (Arts.  32  and  33).     Thus, 

33.  ax  —  6x  +  ex  —  ay  +  by  —  cy 
may  be  written 

(ax  —  bx)  +  (ex  —  a?/)  +  (&»/  -  cy), 
or  (ax  —  6jb  +  ex)  —  (a?/  —  by  +  cy), 

or  (ax  —  a?/)  —  (6x  —  by)  +  (ex  —  cy). 

Whenever  a  factor  is  common  to  every  term  within  a 
parenthesis,  it  may  be  placed  outside  of  the  parenthesis  as 
a  multiplier  of  the  expression  within.     Thus, 

34.  ax8  +  7  —  ex  —  dx2  —  c  +  bx  —  dx9  +  bx2  —  2x 
=  (axs  -  dx3)  +  (bx2  -  dx2)  +  (&x  -  ex  -  2x)  +  (7  -  c) 
=  (a  -  d)xs  +  (b-  d)x2  +  (b  -  c  -  2)x  +  (7  -  c). 

In  this  result,  (a  —  d),  (b  —  d),  (b  —  c  —  2)  are  regarded 
as  the  coefficients  of  x3,  x2,  and  x,  respectively  (Art.  17). 
Hence  we  have  here  placed  together  in  parentheses  the 
coefficients  of  the  different  powers  of  x  so  as  to  have  the 
sign  +  before  each  parenthesis. 

35.  —  a2x  —  la  +  a2y  +  3  —  2x  —  db 

_  -(a2x  -  a2y)  -  (7a  4-  aft)  -  (2x  -  3) 
=  -(X  _  y)a2  -  (7  4-  6)a  -  (2x  -  3). 
We  have  here  placed  together  in  parentheses  the  coeffi- 
cients of  the  different  powers  of  a  so  as  to  have  the  sign  — 
before  each  parenthesis. 

In  the  following  four  examples  place  together  in  paren- 
theses the  coefficients  of  the  different  powers  of  x  so  that 
the  sign  4-  will  be  before  all  the  parentheses. 

36.  ax4  4-  bx2  +  5  +  2bx  —  ox2  4-  2x4  —  3x. 

Ans.  (a  +  2)x4  +  (b  -  5)x2  4-  (26  -  3)x  +  5. 

37.  36x2  —  7  -  2x  4-  «6  4-  5ax3  +  ex  —  Ax2  —  6x3. 
Ans.  (5a  -  b)x3  +  (36  -  4)x2  +  (c  -  2)x  4-  ab  -  7. 

38.  2  —  7x3  +  5ax2  -  2cx  4-  9ax3  +  7x  —  3x2. 

Ans.  (9a  -  7)x3  4-  (5a  -  3)x2  4-  (7  -  2c)x  +  2. 


36  EXAMPLES. 

39.  2CX5  -  Sabx  +  idx  -  3bx*  -  a*x*  +  x\ 

Ans.  (2c  -  a?)x*  +  (1  -  36)aJ*  +  (4d  -  3a6)a;. 
In  the  following  four  examples  place  together  in  paren- 
theses the  coefficients  of  the  different  powers  of  x  so  that  the 
sign  —  will  be  before  all  the  parentheses. 

40.  ax2  +  5.r3  -  a¥  —  2bx*  -  3x2  -  bx\ 

Ans.  -(a2  +  6).t4  -  (26  -  5).r3  -  (3  -  a)x2. 

41.  7.i-3  —  3c*x  —  abx5  +  5aa;  +  7.r5  —  (ibex*. 

Ans.  —  (ab  —  7)x-5  —  (abc  —  l)x3  —  (3c2  —  5a)x. 

42.  ax2  +  a2a:3  —  bx2  —  5  a;2  —  ex8. 

Ans.  —  (c  —  a2)#3  —  (6  +  5  —  «).r. 

43.  362a;4  -  6a;  -  ax*  -  rac4  -  5c2a;  -  7x*. 

Ans.  -(a  +  c  +  7  -  362)«*  -  (6  +  5c2)a;. 
Simplify  the   following   expressions,   and  in  each  result 
place  together  in  parentheses  the  coefficients  of  the  different 
powers  of  oj.     This  is  known  as  re-grouping  the  terms  accord- 
ing to  the  powers  of  x. 

44.  ax8—  2cx—  [6a;2—  \cx—dx—  (6a;3+3ca;2)  \  —  (ex*— bx)~\. 

Ans.  (a  -  b)x3  -  (6  +  2c) a2  -  (6  +  c  +  d)x. 

45.  ax2  -  3^-aa;3  +  36a;  -  4[£c.f8  -  §(ax  -  bx*)]\. 

Ans.  (3a  +  2c)a;3  +  (a  +  Sb)x2  -  (8a  +  9b)x. 


4G. 


x.5_4^4_iri2ax_4  ( 35aj*_9/^_  6aA-|oaJ*l  |. 


Ans.  (66  +  l)x6  -  (a  +2b)x*  -  (2a  +  3c)a;. 
"We  shall  close  this  chapter  with  a  few  examples  in  Addi- 
tion and  Subtraction. 

47.  To  the  sum  of  2a  —  36  —  2c  and  26  —  a  +  7c  add 
the  sum  of  a  —  4c  +  76  and  c  —  66.  Ans.  2a  -f  2c. 

48.  Add  the  sum  of  2y  —  Sy2  and  1  —  5ys  to  the  remainder 
left  when  1  —  2y2  +  y  is  subtracted  from  b>f.    Ans.  —y2  -f  y. 

49.  Take  a;2  —  y2  from  3xy  —  4y2,  and  add  the  remainder 
to  the  sum  of  Axy  —  x2  —  Sy2  and  2x2  -f-  6i/a.         ^w.  7.r/./. 

50.  Add  together  3.t2  —  7x  +  5  and  2a;3  +  5a;  —  3,  and 
diminish  the  result  by  fix*  +  2.  Ans.  2x*  —  2&. 

51.  What  expression  must  be  subtracted  from  3a  —  56  -f  c 
so  as  to  leave  2a  —  46  +  c?  Ans.  a  —  b. 


«*£ 


EXAMPLES.  37 

52.  From  what  expression  must  3a5  +  o&c  —  Gca  be  sub- 
tracted so  as  to  leave  a  remainder  Gca  —  56c?         Ans.  Sab. 

53.  Subtract  3a;3  —  7x  +  1  from  2a;2  —  ox  —  3,  then 
subtract  the  difference  from  zero,  and  add  this  last  result  to 
2x2  —  2x3  -  4.  Ans.  x*  —  2x. 

54.  Subtract  3ar  —  5x  +  1  from  unity,  and  add  5a;2  —  6a; 
to  the  result.  Ans.  2x'2  —  x. 

55.  To  what  expression  must  Tar5  —  6a;2  —  5a;  be  added 
lso  as  to  make  \)x3  —  6x  —  7x2?  Ans.  2x3  —  a;2  —  x. 

56.  From  5a;3  +  3a;  —  1  take  the  sum  of  2a;  —  5  +  7 'x* 
and  3a;2  +  4  —  2a;3  +  x.  Ans.  7x3  —  10a;2. 

&z).  Subtract  3a  —  7a3  +  5a2  from  the  sum  of  2  +  8a2  —  a3 
and  2a3  -  3a2  +  a  -  2.  Ans.  8a3  —  2a. 

58.  If  3x2  —  7a;  +  2  be  subtracted  from  zero,  what  will 
be  the  result?  Ans.  -3a;2  +  7a;  -  2. 

59.'  Subtract  5a;2  +  3a;  —  1  from  2a;3,  and  add  the  result 
to  3a;2  +  3x  -  1.  Ans.  2xs  -  2a;2. 

GO.  What  expression  must  be  added  to  5a;2  —  7a;  +  2  to 
produce  7a;2  -  1?  Ans.  2a;2  +  7a;  -  3. 

61.  What  expression  must  be  added  to  4a;3  —  3a;2  +  2  to 
produce  4a;3  +  7a;  -  6  ?  Ans.  3a;2  +  7a;  -  8. 

62.  What  expression  must  be  subtracted  from  9a;2+lla;— 5 
so  as  to  leave  6a;2  -  17a;  +  3?  Ans.  3a;2  +  28a;  —  8. 

63.  From  what  expression  must  11a2  —  5a&  —  Ibc  be 
subtracted  so  as  to  give  for  remainder  5a2  +  7a&  +  76c  ? 

Ans.  16a2  +  2a6. 


38  MULTIPLICATION — RULE    OF  SIGNS. 


CHAPTER     IV. 

MULTIPLICA  T  I  O  N. 

35.  Multiplication  in  Algebra  is  the  process  of  taking 
any  given  quantity  as  many  times  as  there  are  units  in  any 
given  number.* 

The  Multiplicand  is  the  quantity  to  be  taken  or  multiplied. 

The  Multiplier  is  the  number  by  which  it  is  multiplied. 

The  Product  is  the  result  of  the  operation. 

The  multiplicand  and  multiplier  taken  together  are  called 
Factors  of  the  product. 

In  Algebra  as  in  Arithmetic,  the  product  of  any  number  of 
factors  is  the  same  in  whatever  order  the  factors  may  be  taken 
(Art.  21).  Thus,  2x3x5  =  2x5x3  =  3x5x2, 
and  so  on.     In  like  manner  abc  =  acb  =  bca,  and  so  on. 

Also  2a  x  36  =  2  x  a  x  3  x  & 

=  2x3«XaX& 
=  6a&. 

36.  Rule  of  Signs.  —  The  rule  of  signs,  and  especially 
the  use  of  the  negative  multiplier,  usually  presents  some 
difficulty  to  the  beginner. 

(1)  if  +a  is  to  be  multiplied  by  +  <\  this  indicates 
that  -\-a  is  to  be  taken  as  many  times  as  there  are  units  in 
c  Now  if  +a  be  taken  once,  the  result  is  -\-a  ;  if  it  be 
taken  twice,  the  result  is  evidently  +2a;  if  taken  three 
times,  the  result  is  +  3«;  and  so  on.  Therefore  if  +a  be 
taken  c  times,  it  is  +ca  or  +a<-.     That  is 

+a  x  +c  =  +oc. 

(2)  If  —a  is  to  be  multiplied  by  -+-C,  this  indicates  that 
—a  is  to  be   taken    as  many  times  as   there   arc  units   in  c. 

*  This  •definition  Is  true  only  of  whole  numbers. 


RULE   OF  SIGNS.  39 

Now  if    —a   be  taken   once,   the   result   is   —a;    if   it   be 
taken  twice,  the   result  is  —  2a;   if  taken  three  times,  the 
result  is  —3a;    and  so  on.      Therefore  if  —  a  be    taken   c 
times,  it  is  —ca  or  —ac.     That  is, 
—  a  x   +c  =  —ac. 
Similarly         —  3  x   +4  =  —3  taken  four  times 
=  -3  -3  -3  -3 
=   -12. 

(3)  Suppose  +«  is  to  be  multiplied  by  —  c.  We  have 
illustrated  the  difference  between  +c  and  — c  (Art.  20),  by 
supposing  that  +c  represents  a  line  of  c  units  measured  in 
one  direction,  and  —  c  a  line  of  c  units  measured  in  the  oppo- 
site direction.  Hence  if  +a  is  to  be  multiplied  by  — c,  this 
indicates  that  +  a  is  to  be  taken  as  many  times  as  there  are 
units  in  +c,  and  further  that  the  direction  of  the  line  which 
represents  the  product  is  to  be  reversed. 

Now  +a  taken  +c  times  gives  +ac  ;  and  changing  the 
sign,  which  corresponds  to  a  reversal  of  direction,  we  get 
—ac.     That  is 

+a  x   —  c  =  —ac. 

Similarly  +3  X  —4  indicates  that  3  is  to  be  taken  4  times, 
and  the  sign  changed.  The  first  operation  gives  +12,  and 
the  second  —12.     That  is 

+3  x   -4  =  -12. 

(4)  If  —a  is  to  be  multiplied  by  — c,  this  indicates  that 
—  a  is  to  be  taken  as  many  times  as  there  are  units  in  c,  and 
then  that  the  direction  of  the  line  which  represents  the 
product  is  to  be  reversed. 

Now  —a  taken  c  times  gives  —  ac;  and  changing  the 
sign,  which  corresponds  to  a  reversal  of  direction,  we  get 
-\-ac.     That  is 

—  a  x   —  c  =  +ac. 

Similarly  —3  x  —4  indicates  that  —3  is  to  be  taken  4 
times,  and  the  sign  changed.  The  first  operation  gives  —12, 
and  the  second  +12.     That  is, 

-3  X   -4  =  +12. 


40  EXAMPLES. 

(3)  is  sometimes  expressed  as  follows  :  +«  multiplied  by 
— c  iudicates  that  +a  is  to  be  taken  as  many  times  as  there 
are  units  in  c,  and  then  the  result  subtracted.  Now  +<i 
taken  +c  times  gives  -foe,  and  changing  the  sign,  in  order 
to  subtract  (Art.  29),  we  get  —  ac. 

Similarly  (4)  indicates  that  —a  is  to  be  taken  as  many 
times  as  there  are  units  in  c,  and  the  result  subtracted. 
Now  —a  taken  +c  times  gives  — ac,  and  changing  the  sign, 
in  order  to  subtract,  we  get  -f-«c. 

Hence  we  have  the  following  Rule  of  Signs:  The  product 
of  two  terms  with  like  signs  is  +  ;  the  product  of  tico  terms 
with  unlike  signs  is  — . 

To  familiarize  the  beginner  with  the  rule  of  signs  we  add 
a  few  examples  in  substitution,  where  some  of  the  symbols 
denote  negative  quantities. 


EXAMPLES. 

If   a  =  -2,  b  =  3,  c  =  -1,  x  =  -5,  y  —  4,  find  the 
value  of  the  following  : 

1.  3a°-b  =  3(-2)2  X3  =  3x4x3  =  36. 

2.  -7a*bc  =  -7(-2)3  x  3  x  (-1)  =  -7  x  -  8  x  3 
X  -1  =  -1G8. 

Ans.  -1000. 

375. 

500. 

-224. 

-10. 

-9. 

40. 

=  4,  y  =  1, 

Ana.  10. 

lis. 
-130. 


3.    8a6c2.         A\ 

w.  —48. 

11. 

8cV. 

4.    Ga2c2. 

24. 

12. 

8c8a?». 

5.    -2a4to. 

480. 

13. 

4c5.r3. 

G.    5a V2. 

500. 

14. 

7a6c4. 

7.    —lc4:nj. 

140. 

15. 

-4aV. 

8.    -Saxs. 

-2000. 

16. 

-62c2. 

9.    -5aW. 

-180. 

17. 

2a2c8a?. 

10.   -7a8c8. 

-5G. 

If  a  =  -4,  b  = 

=  -3,  c  = 

-1,/ 

=  0,  a; 

id  the  value  of 

18.    3a2  +  bx  - 

4q/. 

19.    fr  -  2b3  - 

ex8. 

20.    3«V  -  56% 

!  -  2c8. 

THE  MULTIPLICATION   OF  MONOMIALS.  41 

21.  2a3  —  36s  +  7c?/4.  Ans.  —  54. 

22.  36V  -  4^7  -  6<'4x-  3* 

23.  2\f(ac)  -  3\f(xy)  +  V^c4).  1. 
It  is  convenient  to  make  three  cases  in    Multiplication, 

(1)  the  multiplication  of  monomials,  (2)  the  multiplication 
of  &  polynomial  by  a  monomial,  and  (3)  the  multiplication  of 
■polynomials. 

37.  The  Multiplication  of  Monomials.  —  Since  by 
definition  (Art.  12)  we  have 

a4  =  aaaa, 
and  a6  =  aaaaaa, 

.'.     a4  x  aG  =  aaaaaaaaaa 
=  a10. 

Also  3a2  =  3aa, 

7a3  =  7aaa. 
3a2  x  7a3  =  3  x  7  x  aaaaa 
=  21a5. 
Similarly    5«3£r  X  6a2b*xz  =  5aaabb  X  Gaabbbbxx 

=  30a5b*x2. 
Also    4a3c2  x  3c3a;2  x  3a;2  =  Aaaacc  x  3ccca;a;  x  3x-a; 

=  3Ga3cbx\ 
Hence  for  the  multiplication  of  monomials  we  have  the 

following 

Rule. 

Multiply  together  the  numerical  coefficients,  annex  to  the 
result  all  the  letters,  and  give  to  each  letter  an  exponent  equal 
to  the  sum  of  its  exponents  in  the  factors. 

For  example  2a;2  X  3a;4  X  a;6  =  Ga:2+4  +  6  =  Gx12. 

Also      5a263  x  262c4  x  3c2cZ4  =  30aWd4. 

Note.  —  The  beginner  must  be  careful,  in  applying  this  rule,  to 
observe  that  the  exponents  of  one  letter  cannot  combine  in  any  way 
with  those  of  another.  Thus,  the  expression  4a2b3c4  admits  of  no 
further  simplification. 

The  product  of  three  or  more  expressions  is  called  the 
continued  product. 


42  EXAMPLES. 

38.   To   Multiply  a  Polynomial  by  a  Monomial 

—  Suppose  we  have  to  multiply  (a  +  b)  by  3  ;   that  is,  take 
a  +  b  3  times.     We  have 

3  («  +  6)  =  (a  +  b)  +  (a  +  6)  +  (a  +  &) 
=  (a  +  a  +  a  taken  three  times) 
together  with  (6  +  b  +  b  taken  three  times) 

=  3a  +  ob. 
Similarly  7  (a  +  6)  =  7a  +  76. 

m(a  +  b)  =  (a  +  a  +  a  + taken  m  times) 

together  with  (b  +  b  +  b  + taken  ?/i  times) 

=  ma  +  ?n& (1) 

Also  m(a  —  6)  =  (a  +  a  +  a  + taken  m  times) 

together  with  {  —  b  —  b  —  b— taken  m  times) 

=  ma  —  mb (2) 

Similarly 

m(a  —  6  +  c)  =  7?ia  —  ?ji6  +  ?wc. 
This  is  generally  called  the  Distributive  Law. 
Hence,  to  multiply  a  polynomial  by  a  monomial,  we  have 
the  following 

Rule. 

Multiply  each  term  of  the  polynomial   separately  by  the 
monomial,  and  collect  the  results  to  form  the  complete  product. 
For  example, 

-i(.ir  +  2xy  -  4z)  =  4x2  +  8xy  -  16z, 
(4a2  -  ly  -  Hz3)  x  '3xy2  =  12afy2  -  21a;?/8  -  24a#%8. 

(|a2    _    1(^    _    &S)    x     gaS63    _    4a4&2    _    a868    _    ga^4. 

EXAM  PLES. 

Multiply  together 

1.  4a-b3  and  7a5.  -4ws.  28a'68. 

2.  3a46V  and  5a86a?.  15a768ai*. 

3.  2x2yz3  and  x5>fz.  2ajTyV. 

4.  a6  +  6c  and  a36.  a*6s  +  a*b*c. 

5.  5a  +  ."•//  and  2a;-.  10.x3  +  6afy. 
0.  6c  +  ca  —  a6  and  a6c.  a6V  4-  «~b<:~  —  "-/'■'■. 
7.    5xhf  +  a*/9  -  7.('-//-and  S.ry\  40«V  +  8«Y  -  56*Y. 


8. 

6a?bc  - 

7a&2c2  and  <rb'\ 

.4ws.  Ga563c 

9. 

a*b8x  - 

56a  and  2a8a>6. 

2a'b3x6  - 

.0. 

8a268  - 

§&2c3  and  §a&4. 

12a3i 

THE    MULTIPLICATION   OF  POLYNOMIALS.  43 

-   7rt364C2. 

lOaW5. 
—  a66c8. 

39.  The  Multiplication  of  a  Polynomial  by  a 
Polynomial.  —  Suppose  we  have  to  multiply  c  +  d  by 
a  +  &. 

Here  we  are  required  to  take  c  +  d  as  many  times  as 
there  are  units  in  a  +  &,  i.e.,  we  are  to  take  c  +  d  as  many 
times  as  there  are  units  in  a,  and  then  add  to  this  product 
c  +  d  taken  as  many  times  as  there  are  units  in  b. 

Hence  (a+&)  (c+d)  =  (c+d)  taken  a  times 
together  with  (e+d)  taken  &  times 

=  (c+d)a+(c+d)& 
=  ac+ad+&c+5d[(l)ofArt.38].  (1) 
Again  (a— 6)  (c+d)  =  (c+d)  taken  a  times 
diminished  by  (c+d)  taken  &  times 

=  (c+d)a—  (c+d)b 
=  ftc+ad-  (bc+bd)  [(l)of  Art.  38] 
=  ac+ad-&c-6d  (Art.  33) .  .     .   (2) 
Also     (a + 6 )  (c — d)  =  (c — d)  taken  a  times 
together  with  (c— d)  taken  &  times 

=  (c— d)a+(c— d)6 
=  ac-ad+6c-6d[(2)ofArt.38].  (3) 
Lastly  («  —  6)  (c— d)  =  (c—d)  taken  a  times 
diminished  by  (c~ d)  taken  6  times 

=  (c— d)a—  (c— d)& 
=  ac'—ad-  (bc—bd)  [(2)  of  Art.  38] 
=  oc-od-6c+6d  (Art.  33).  .     .   (4) 
Hence,  to  multiply  one  polynomial  by  another,  we  have 
the  following 

Rule. 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the 
multiplier;  if  the  terms  multiplied  together  have  the  same  sign, 
prefix  the  sign  +  to  the  product,  if  unlike,  prefix  the  sign  —  ; 
then  add  the  partial  products  to  form  the  complete  product. 


44  EXAMPLES. 

If  we  consider  each  term  in  the  second  member  of  (4), 
and  the  way  it  was  produced,  we  find  that 

+a  x  +c  =  +ae. 

-fa  x  —  d  —  —  ad. 

-b  x  -fc  =  -6c. 

_6  x  -d  =  +bd. 
These  results  enable  us  again  to  state  the  rule  of  signs, 
and  furnish  us  with  another  proof  of  that  rule,  in  addition 
to  the  one  given  in  Art.  36.  This  proof  of  the  rule  of  signs 
is  perhaps  a  little  more  satisfactory  than  the  one  given  in 
Art.  36,  though  it  is  not  quite  so  simple. 

EXAMPLES. 

1.    Multiply  x  +  7  by  x  -f-  5. 
The  product  =  (x    +  7)  (x  +  5) 

=   x'2  +  Ix  +  5x  +  35 
=  x2  +  12b  +  35. 
Rem.  —  It  is  more  convenient  to  write  the  multiplier  under  the 
multiplicand,  and  begin  on  the  left  and  work  to  the  right,  placing 
like  terms  of  the  partial  products  in  the  same  vertical  column,  as 

follows : 

x  +    7 
x   +    5 


addition 

x-  +    7x 
+    5x  +  35 

by 

x*  +  12x  +  35. 

2. 

Multiply 

3x 

—  iy  by  2x  —  3y. 
3x    —     Ay 
2x    -    Sy 

Gx2  -    8xy 

-    9xy  +  12?/2 

by  addition  6x*  —  11  xy  +  12jr. 

Here  the  first  line  under  the  multiplier  is  the  product  of 
the  multiplicand  by  2x  ;  the  second  line  is  the  product  of  the 


EXAMPLES.  45 

multiplicand  by  —3?/;    like   terms  are  placed  in  the   same 
vertical  column  to  facilitate  addition. 
Find  the  product  of  the  following  : 


3. 

a;  —  7  and  x  —  10. 

Ans.  x2  —  17a;  +  70. 

4. 

x  —  7  and  x  +  10. 

a;2  +  3a;  —  70. 

5. 

aj  —  12  and  05  —  1. 

x'  -  13x  +  12. 

6. 

x  —  15  and  x  -\-  15. 

a;2  -  225. 

7. 

— .c  -  2  and  —a;  —  3. 

a;2  +  5a;  +  6. 

8. 

—x  +  5  aud  —  x  —  5. 

x2  -  25. 

9. 

a;  —  17  aud  %  +  18. 

a;2  +  x  -  306. 

.0. 

—x  —  16  and  — flJ  +  16. 

a;2  -  256. 

.1. 

2x  —  3  and  x  +  8. 

2a;2  +  13a;  -  24. 

.2. 

3a;  -  5  and  2aj  +  7. 

6a;2  +  11a;  -  35. 

.3. 

4aa  -  5a6  +  662  and  2a2  - 

-  Sab  +  463. 

4a2  -    5a6    +    6b2 
2«2  -    3a6    +    4&2 

8a4  -  10a36  +  V>a2b2 

-  12a86  +  15a262  -  18«&3 

+  i6a262  -  20a63  +  2464. 

8a4  -  22a36  +  43a262  -  38o63  +  2464. 


Here  the  first  line  under  the  multiplier  is  the  product  of 
the  multiplicand  by  2a2 ;  the  second  line  is  the  product 
by  —  3ab ;  the  third  by  4&2;  like  terms  are  set  down  in  the 
same  vertical  column  to  facilitate  the  addition. 

The  student  will  observe  that  both  the  multiplicand  and 
multiplier  are  arranged  according  to  the  descending  powers 
of  a  (Art.  19).  Both  factors  might  have  been  arranged 
according  to  the  ascending  powers  of  a.  It  is  of  no  conse- 
quence which  order  we  adopt,  but  we  should  take  the  same 
order  for  the  multiplicand  and  multiplier. 

If  the  multiplier  and  multiplicand  are  not  arranged  accord- 
ing to  the  powers  of  some  common  letter,  it  will  be  convenient 
to  rearrange  them.     Thus  : 


46  EXAMPLES. 

14.  Multiply  3.r  -f  4  +  2a;2  by  4  +  2x2  —  Bx.  Arranging 
the  factors  according  to  the  descending  powers  of  x,  the 
operation  is  as  follows  : 

2x-  +  Bx    -f  4 

2a;2  -  3a;    +  4 


4a;4  +  6a;3  +  8a;2 

-  6x3  -  9x*  -  12a; 

+  8.r  +  \2x  +  16 

4x4  +  7.c2  +  16. 

15.    Multiply  a2  4-  b2  +  c2  —  a&  —  be  —  ca  by  a ■  +  6  +  c. 

Arrange  according  to  descending  powers  of  a. 
a2  —  a&    —  ac    4-  &2    —    be    4-  c2 
a   4-6     +  c 


a3  —  a26  —  a2c  4-  o&2  —    a&c  4-  ac2 

+  «'26  —  ab2  —    abc  4-  b3  —  b2c  +  6c2 

4-  a2c  —    abc  —  ac2  +  62c  —  6c2  4  c3 

a3  -  3a&c  +  bz  +  e8 

Rem.  —  The  student  should  notice  that  he  can  make  two  exercises 
in  multiplication  from  every  example  in  which  the  multiplicand  and 
multiplier  are  different  polynomials,  by  changing  the  original  multi- 
plier into  the  multiplicand,  and  the  original  multiplicand  into  the 
multiplier.  The  result  obtained  should  be  the  same  in  both  opera- 
tions. The  student  can  therefore  test  the  correctness  of  his  work  by 
interchanging  the  multiplicand  and  multiplier. 

Multiply  together 

16.  a2  —  ab  +  b2  and  a2  4-  ab  +  b2.     Ans.  a*  4  a2b2  4  o4. 

17.  x2  4-  By1  and  x  4  4y.  x3  4  Ax2:/  4-  oxy2  4  12//?. 

18.  a;4  —  x2y2  4  ?/4  and  x2  +  y2. 

19.  a2  —  2ax  4-  Ax2  and  a2  +  2ax  +  4 x2. 

20.  16a2  +  12ab  +  9&2  and  4a  -  36. 

21.  a2x  —  ax2  4  x3  —  a3  and  &  +  Q» 

22.  2a;3  -  3a;2  +  2sc  and  2a;2  +  3a;  +  - 

23.  -a5  4-  «4&  -  a3b2  and  -a  -  6.  a6  +  «8&8- 

24.  a3  4  2as6  4  2a&2  and  aa  -  2a&  4  26s.       a5  4-  4a64. 


a;6  +  yr'. 

a4  4-  4a2a;2  +  16a;4. 

64a3  -  276". 

a;4  -  a4. 

4a;5  -  x3  4   !•'•■ 

EXAMPLES.  47 

When  the  coefficients  are  fractional  we  use  the  ordinary 
process  of  multiplication,  combining  the  fractional  coefficients 
by  the  rules  of  Arithmetic* 

25.    Multiply  4a2  -  4a&  +  f&2  by  4a  +  4> 

4a2  -     fa&    +  f&2 
4a    +    lb 


la3   _        la2ft    _|_    la&2 


lfl3    _    _5^    +    la&2    +    |&8< 

Multiply  together 

26.  |a2  +  4a  +  i  and  A/t  —  ^.      J.?is.  |a8  +  ^a  —  ^. 

27.  f«2  +  a^  +  \tf  and  4a  -  |#.  -fa3  -  \tf. 

28.  4s2  -  fe  -  f  and  4s2  4-  f  a  -  |.     K  -  ffa;2  +  tV 

29.  fax  +  fa2  +  4a2  and  fa2  +  fa;2  -  faa?.        ±a4  +  a;4. 

It  is  sometimes  desirable  to  indicate  the  product  of  poly- 
nomials, by  enclosing  each  of  the  factors  in  a  parenthesis, 
and  writing  them  in  succession.  When  the  indicated  multi- 
plication has  been  actually  performed,  the  expression  is  said 
to  be  expanded,  or  developed. 

Expand  the  following : 

30.  (2a  +  36)  (a  —  6).  Ans.  2a2  +  ab  —  3b2. 

31.  (a2  +  ax  +  a;2)  (a2  —  ax  +  x2).  xA  +  a2x2  +  a*. 

32.  (a2  +  2a6  +  2&2)  (a2  -  2a&  +  26s).  «4  +  4&4. 

33.  [x  -  3)(x  +  4)  (a;  —  5)(a>  +  (5). 

-4ns.  a;4  +  2x3  -  41a?  -  42a;  +  3G0. 

34.  (a2  +  ab  +  &2)  (a3  -  a26  +  &3)  (a  -6). 

^?is.  a6  -  a6&  +  a2&4  -  &6. 

35.  (2a?  +  4x2  +  8a;  4-  16)  (3a;  -  6).  6a;4  -  96. 

36.  (x*  -{-  x2  +  x  -  \){x  -  \).  xi  -  2x  +  1. 

37.  (x  +  a)  [(a;  +  b)(x  +  c)  -  (a  +  6  4-  c)(a;  +  6) 
+  (a2  4-  ab  +  ft2)].  -4ws.  a;3  +  a3. 

*  The  student  is  supposed  to  be  familiar  with  Arithmetic  fractions,  which  are  the 
only  fractions  that  are  used  in  this  work  previous  to  Chapter  VIII. 


48       MULTIPLICATION  BY  INSPECTION — EXAMPLES. 

40.  Multiplication  by  Inspection.  —  Although  the 
result  of  multiplying  together  two  binomial  factors  can 
always  be  obtained  by  the  methods  explained  in  Art.  3(J, 
yet  it  is  very  important  that  the  student  should  learn  to 
write  down  the  product  rapidly  by  inspection.  This  is  done 
by  observing  in  what  way  the  coefficients  of  the  terms  in  the 
product  arise ;  thus 

(x  +  5)  (a;  +  3)  =  a2  -f  5a  +  3x  +  15 

=  x2  +  8x  +  15. 
(x  -  5)  (x  +  3)  =  x2  -  bx  +  3a;  -  15 

_  xi  _  2x  -  15. 
(x  +  5)(aj  -  3)  =  a-2  +  5x  -  3a;  -  15 

=  x2  +  2x  —  lf>. 
(x  -  5)(cc  -  3)  =  x2  -  5x  -  Sx  +  15 
=  x2  -  8x  +  15. 
It  will  be  noticed  in  each  of  these  results  that : 

1.  The  product  consists  of  three  terms. 

2.  The  first  term  is  a;2,  and  the  last  term  is  the  product  of 
the  second  terms  of  the  two  binomial  factors. 

3.  The  middle  term  has  for  its  coefficient  the  Algebraic 
sum  of  the  second  terms  of  the  two  binomial  factors. 

Hence  the  intermediate  steps  in  the  work  may  be  omitted, 

and  the  product  written  down  at  once,  as  follows : 

(x  +  2)  (x  +  3)  =  a;2  +  5as  +    6. 

(x  -  3)  (a;  +  4)  =  x2  +    x  -  12. 

(x  +  6)  (a;  -  9)  =  x2  -  3a;  -  54. 

(x  -  4y)(x  -  lOy)  =  x2  -  Uxy  +  40y2. 

(as  -  by)  \x  +    Gy)  =  x2  +      xy  -  30t/2. 

EXAMPLES. 

Write  down  the  values  of  the  following  products, 

l.    (a:  +  8)  (x  -  5).  Ana.  x*  +  3aj  -  40. 

•_'.    (./•  -  3) (x  +  10).  x2  +  7a?  -  30. 

3.    (.c  +  7)  (a;  -  !)).  x2  -  2x  -  63. 

I.    (r  -  4) (a;  +  11).  x2  +  7as  -  44. 


SPECIAL   FORMS   OF  MULTIPLICATION.  49 


5. 

(»+  2) (a-  5). 

Ans 

.  x2  -  Sx  -  10. 

6. 

(*+  9)(tB-  5). 

x2  +  4x  —  45. 

7. 

(»  -8)(oj  +  4). 

a-a  _  4x  _  32. 

8. 

(x  -  G)(x  +  13). 

a;'2  +  7x  —  78. 

9. 

0  -  11) 0+  12). 

x-2  +  x  -  132. 

10. 

(x  —  3a)  (x  +  2a) . 

#2  —  a#  —  6a2. 

11. 

(x  -  9b)(x  +  86). 

x2  -  bx  -  72&2. 

12. 

(x  -  ly)  (x  -  8y) . 

X2 

—  15xy  +  562/2. 

43.  Special  Forms  of  Multiplication  —  Formulae. 

—  There  are  some  examples  iu  multiplication  which  occur 
so  often  in  Algebraic  operations  that  they  deserve  especial 
notice. 

If  we  multiply  a  +  b  by  a  +  b  we  get 

(a  +  b)  (a  +  b)    =  a2  +  2a6  +  b2 ; 
that  is  (a  +  b)2  =  a2  +  2ab  +  b2.  .     .     .   (1) 

Thus  £/*e  square  of  the  sum  of  two  numbers  is  equal  to  the 
sum  of  the  squares  of  the  two  numbers  increased  by  twice  their 
product. 

Similarly,  if  we  multiply  a  —  b  by  a  —  b  we  get 

(a  -  b)2  =  a2  -  2ab  +  b2 (2) 

Thus  the  square  of  the  difference  of  two  numbers  is  equal 
to  the  sum  of  the  squares  of  the  two  numbers  diminished  by 
twice  their  product. 

Also,  if  we  multiply  a  +  b  by  a  —  b  we  get 

(a  +  b)  (a  -  b)  =  a2  -  b2 (3) 

Thus  the  product  of  the  sum  and  difference  of  two  numbers 
is  equal  to  the  difference  of  their  squares. 

Because  the  product  of  two  negative  factors  is  positive 
(Art.  36),  it  follows  that  the  square  of  a  negative  number 
is  positive.     For  example, 

(-a)2  =a*  =  (+a)2, 
and  (b  —  a)2  =  a2  —  2ab  +  b2  =  (a  —  b)2. 


50  EXAMPLES. 

Hence  ft2  —  2ab  -f  b2  is  the  square  of  both  a  —  b  and 
b  —  a. 

Rem.  1.  — Equations  (1),  (2),  and  (3)  furnish  simple  examples  of 
one  of  the  uses  of  Algebra,  which  is  to  prove  general  theorems 
respecting  numbers,  and  also  to  express  those  theorems  briefly. 

For  example,  the  result  (a  -+-  b)(a  —  b)  =  a2  —  b2  is 
proved  to  be  true,  and  is  expressed  thus  by  symbols  more 
compactly  than  it  could  be  by  words. 

A  general  result  thus  expressed  by  symbols  is  called  a 
formula;  hence  a  formula  is  an  Algebraic  expression  of  a 
general  rule. 

Rem.  2.  — We  may  here  indicate  the  meaning  of  the  sign  ±  which 
is  made  by  combining  the  signs  +  and  — ,  and  which  is  called  the 
double  sign. 

By  using  the  double  sign  we  may  express  (1)  and  (2)  in 
one  formula  thus : 

(a  ±  b)2  =  a2  ±  2ab  +  b2,  .     .     .     .   (1) 

where  ± ,  read  plus  or  minus,  indicates  that  we  may  take 
the  sign  +  or  — ,  keeping  throughout  the  upper  sign  or  the 
lower  sign.  Formulae  (1),  (2),  and  (8)  arc  true  whatever 
may  be  the  values  of  a  and  b. 

The  following  examples  will  illustrate  the  use  that  can  be 
made  of  formulae  (1),  (2),  and  (.'5).  The  formulae  will 
sometimes  be  of  use  in  Arithmetic  calculations.     Thus 


EXAMPLES. 

1.  Required  the  difference  of  the  squares  of  127  and  L23. 
By  formula  (3)  we  have 

(127)2  -  (128)2  =   (127  +  128)  (127  -   128) 
=     2;")0  x    1  =  1(«)0. 

2.  Required  the  square  of  29. 
By  formula  (2) 

(2!»)-=  (30  -  l)9  =  900  -  GO  +  1  =  841. 


EXAMPLES.  51 

3.  Required  the  product  of  53  by  47. 
By  formula  (3) 

53  X  47  =  (50  +  3)  (50  -  3)  =  (50)2  -  (3)2 
=    2500  -  9  =  2491. 

4.  Required  the  square  of  34. 
By  formula  (1) 

(34)2  =  (30  +  4)2  =  900  +  240  +  1G  =  1156. 

5.  Required  the  square  of  Ax  +  oy. 

We  can  of  course  obtain  the  square  by  multiplying  Ax  +  3y 
by  itself  in  the  ordinary  way.  But  we  can  obtain  it  by 
formula  (1)  more  easily,  by  putting  Ax  for  a  and  3y  for  b. 
Thus 

(4a;  +  3y)2  =  (Ax)*  +  2(4a%)  +  (3yY 
=  16x-2    +  2Axy        +  9t/2. 

G.    Required  the  square  of  x  +  y  +  %• 
Denote  x  +  y  by  a ;  then  x-\-y  +  z  =  a-\-z;  and  by 
(1)  we  have 

(a  +  z)2  =  a2  +  2az  -f-  z2 

=  (a>  +  yY  +  2(x  +  y)g  +  22 
=  x2  +  2xy  +  ^  +  2xz  +  2tf2  +  z2 
Thus  (»  +  y  +  z)2  =  f2  -f-  r     +  23  +  2an/  +  2yz  +  2xz. 

That  is,  the  square  of  the  sum  of  three  numbers  is  equal 
to  the  sum  of  the  squares  of  the  three  numbers  increased  by 
twice  the  products  of  the  three  numbers  taken  two  and  two. 
7.   Required  the  square  of  p  —  q  +  r  —  s. 

Deuote  p—q  by  a  and  r—s  by  b  ;  then  p  — q+r— s=a+b  ; 
and  by  (1)  we  have 

(a+&)2=aa+2a&+&2=  (p-qy+2(p-q)  (r-s)  +  (r-s)2. 

Then  by  (2)  we  expand  (p  —  qY  and  (/•  —  s)'2. 

Thus  (p  —  q  +  r  —  s)2 

=  p2  —  2pq  +  g2  +  2  (jpr  —  ps  —  qr  +  </s)  +  r2—  2rs  +  s2 
=  p'2  +  2a  +  >"  +  s9  +  2pr  +  2gs  —  2pg  —  2ps  —  2qr  —  2rs. 


52  EXAMPLES. 

8.  Required  the  product  of  p  —  q+r—s  and  p— q— r+s. 
Let  p  —  q  —  a  and  r—s  =  b  ;  then  p  —  q  +  r  —  s  =  a  +  6, 

and  p  —  g  —  r  +  s  =  a  —  6;  and  by  (3)  we  have 

(a+6)  (a-6)  =a2-62=  (p-g)2-  (r-s)2 

=  p2-2pg+g2-(V2-2rs+s2)  by  (2). 

Thus(p— tf+r— s)(p— q  —  r+s)=p2+g2— r2— s2— 2pq+2rs. 

From  these  examples  we  see  that  by  using  formula?  (1), 
(2),  and  (3),  the  process  of  multiplication  may  be  often 
simplified.  The  student  is  advised  first  to  go  through  the 
work  fully  as  we  have  done ;  but  when  he  becomes  more 
familiar  with  this  subject,  he  may  dispense  with  some  of  the' 
work,  and  thereb}'  simplify  the  multiplication  still  more. 
Thus  in  the  last  example  he  need  not  substitute  a  and  &,  but 
apply  formula  (3)  at  once,  and  then  (2),  as  follows : 

[(P  -  <?)  +  (r  -  .)]  [(f>  -  q)  -  (r  -  «)]  =  (p  -  </)2 

—  (r  —  s)2  =  p2  —  2p§  +  <?2  —  ?'2  +  2rs  —  s2. 

9.  Required   the    product    of    a  +  b  +  c,    a  +  &  —  c, 
a  —  &  +  c,  b  +  c  —  a. 

By  (3)  and  (1)  we  obtain  for  the  product  of  the  first  two 
factors, 

(a  +  6  +  c)  (a  +  &  -  c)  =  2ao  -f  a8  +  &2  -  c2.  .   (1) 
By  (3)  and  (2)  we  obtain  for  the  product  of  the  last  two 
factors, 

(a  _  &  +  c)  (b  +  c-a)  =  2ab  -  (a2  +  &8  -  c2).  .   (2) 
Multiplying  together  (1)  and  (2),  we  obtain 
(2a&)2  -  (a2  +  b2  -  c2)2 

=  2a2&2  +  2&2c2  +  2o2c2  -  a4  -  b*  -  c4.  .     .     .   (3) 

Solve  the  following  examples  in  multiplication  by  formulae 
(1),  (2),  and  (3). 

10.  (15a  4-  U.y)2.  Ans.  225a3  +  420a-?/  +  196?/2. 

11.  (7a;2  -  r>y2)2.  49a;4  -  70aV  +  25y4. 

12.  (a;2  +  2x  -  2)2.  a;4  +  4a;3  -  8x  +  4. 

13.  (a;2  -  bx  +  7)2.  a;4  -  10a;3  +  39a;2  -  70a  +  49. 


IMPORTANT  RESULTS  IN  MULTIPLICATION.  53 

14.  (2x2-3x-4:)2.         Ans.  4x4- 12a3  -  7a;2  + 24.K+ 16. 

15.  (x  +  2y  +  3z)2.         x2  +  4y2  +  9z2  +  Axy  +  6a-z  -f  12yz. 

16.  (aja  +  a^  +  y2)(»2  +  a^-y2).       jb*  +  2a?y  +  ary  - &. 

17.  (x2  +  an/  +  y2)  (a?  -  any  +  r) .  a?*  +  x2y2  +  y\ 

42.  Important  Results  in  Multiplication.  —  There 
are  other  results  in  multiplication  which  are  important, 
although  they  are  not  so  much  so  as  the  three  formulae  in 
Art.  41.  We  place  them  here  in  order  that  the  student  may 
be  able  to  refer  to  them  when  they  are  wanted  ;  they  can  be 
easily  verified  by  actual  multiplication. 

(a  +  &)(a2  -  ab  +  b2)  =  a3  +  b3.  .     .     .   (1) 

(a  -  b)  (a2  +  «&  +  &2)  =  «3  -  b3.  .     .     .   (2) 

(a  +  6)3  =  (a+&)  (a2  +  2ab  +  &2)  =  a3  +  3a2&  +  3a&2  +  &3.   (3) 

(a- 6)8=  (a- 6)  (a2  -  2ab  +  &2)  =  a3  -  3a2&  +  3a&2  -  Z>3.    (4) 

(a  +  b  +  c)3  =  a3  +  3a2(&  +  c)  +  3a(6  +  c)2  +  (6  +  c)3 

=a8+68+c8+3a2(&+c)  +3&2(a+c)  +3c2(a+&)  +6abc.   (5) 

Kem.  —  It  is  a  useful  exercise  iu  multiplication  for  the  student  to 
show  that  two  expressions  agree  in  giving  the  same  result.  For 
example,  show  that 

(a  —  b)(b-  c){c  -  a)  =  a2(c  -  6)  +  b2(a  -  c)  +  c2(6  -  a). 
Here  we  proceed  as  follows:  Multiplying  (a  —  b)  by  (6  —  c)  we 
obtain 

(a  —  6)(6  —  c)  =  ab  —  b2  —  ac  +  be, 

then  multiplying  this  equation  by  c  —  a  we  obtain 
(a  —  b)(b  —  c)(c  —  a)  =  cab  —  c&2  —  ac2  +  be2  —  a2&  +  a&2  +  a2c  —  «6c 
-  a2(c  —  8)  +  &2(a  —  c)  +  c-(6  —  a)  .    .     .   (0) 
Show  that  (a  -  6)2  +  (b  —  c)2  +  (c  -  a)2 

=  2(c  -  b)(c  -  a)  +  2(6  -  a)(b  -  c)  +  2(a  -  6)(«  -  c). 
By  (2)  of  Art.  41  we  obtain 
(a  -  b)2  +  (b-  c)2  +  (c  -  a)2 

=  a2  —  2ab  +  62  +  62  —  2bc  +  c2  +  c2  -  2ac  +  a2 

=  2(o2  +  b2  +  c2  —  a6  —  6c  —  co) (7) 

Now  (c  —  6)(e  —  a)  =  c2  —  ca  —  cb  +  a6, 

(6  —  a)(b  —  c)  —  62  —  be  —  ab  +  ac, 
(a  —  6)(a  —  c)  =  a2  —  ac  —  ab  +  6c; 


54    RESULTS    OF  MULTIPLYING    ALGEBRAIC  EXPRESSIONS. 

therefore,  by  adding  these  three  equations,  we  obtain 
(c  _  6)(c  -  a)  +  (b  -  a)(b  -  c)  +  (a  -  b)(a  -  <■) 

—  a2  +  b'2  +  c2  —  ab  —  ac  —  be;    (8) 
therefore,  from  (7)  and  (S)  we  have 
(a  -  b)2  +  (b  _  c)2  +  (c  -  a)2 

=  2(c  -  6){c  -  a)  +  2(6  -  a)(6  -  c)  +  2(a  -  b)(a  -  c).     (9) 

43.  Results  of  Multiplying  Algebraic  Expressions. 

—  From  an  examination  of  the  examples  in  multiplication, 
the  student  will  recognize  the  truth  of  the  following  laws 
with  respect  to  the  result  of  multiplying  Algebraic  expres- 
sions. 

(1)  In  the  multiplication  of  two  polynomials,  when  the 
partial  products  do  not  contain  like  terms,  the  xohole  number 
of  terms  in  the  final  product  will  be  equal  to  the  product 
of  the  number  of  terms  in  the  multiplicand  by  the  number  of 
terms  in  the  multiplier,  but  will  be  less  if  the  2wrtial  products 
contain  like  terms,  owing  to  the  simplification  produced  by 
collecting  these  like  terms. 

Thus  as  we  see  in  Ex.  17,  Art.  39,  there  are  two  terms  in 
the  multiplicand  and  two  in  the  multiplier,  and  four  in  the 
product,  while  in  Ex.  13  there  are  three  terms  iu  the  multi- 
plicand and  three  in  the  multiplier,  and  only  five  in  the 
product. 

(2)  Among  the  terms  of  the  product  there  are  always  two 
that  are  unlike  any  other  terms;  these  are,  tliat  term  which  is 
the  product  of  the  tioo  terms  in  the  factors  which  contain 
the  highest  power  of  the  same  letter,  and  that  term  which  is  the 
product  of  the  two  terms  in  the  factors  which  contain  the  loio- 
est  poiver  of  the  same  letter. 

Thus  in  Ex.  13,  Art.  39,  there  are  the  terms  8a4  and  2464, 
and  these  are  unlike  any  other  terms  ;  in  fact,  the  other 
terms  contain  a  raised  to  some  power  less  than  the  fourth, 
and  thus  they  differ  from  8a4 ;  and  they  also  contain  a  to 
some  power,  and  thus  they  differ  from  2464. 

(3)  Wlwn  the  multiplicand,  and  multiplier  are  both  homo- 
geneous (Art.  18)  the  product  is  homogeneous,  and  the  degree 


EXAMPLES.  55 

of  the  product  is  the  sum  of  the  numbers  which  express  the 
degrees  of  the  multiplicand  and  multiplier. 

Thus  in  Ex.  13,  Art.  39,  the  multiplicand  and  multiplier 
are  each  homogeneous  and  of  the  second  degree,  and  the 
product  is  homogeneous  and  of  the  fourth  degree.  In  Ex. 
15,  Art.  39,  the  multiplicand  is  homogeneous  and  of  the 
second  degree,  and  the  multiplier  is  homogeneous  and  of  the 
first  degree ;  the  product  is  homogeneous  and  of  the  third 
degree.  This  law  is  of  great  importance,  as  it  serves  to  test 
the  accuracy  of  Algebraic  work  ;  the  student  is  therefore 
recommended  to  pay  great  attention  to  the  degree  of  the 
terms  in  the  results  which  he  obtains. 

EXAMPLES. 

Multiply 

1.  4cr  -  36  by  3a6.  Ans.  12asb  -  9a62. 

2.  8a2'  -  9a6  by  3a2.  24a4  -  27a86. 

3.  3a>2  -  Ay2  +  5z2  by  2x2y.        6x4y  -  Sxhf  +  10x2yz2. 

4.  x2y3  —  y3zA  +  z*x2  by  x2y2z2.        x4y5z2  —  x2y5z6  +  x^yh*. 

5.  2xy2z3  +  3x2y3z  —  bx3yz2  by  2xy2z. 

Ans.  4x2y4z4  +  Cxsy5z2  —  10xYzs. 

6.  -2a26  -  4a62by  -7a262.    '  14a463  +  28a364. 

7.  8xyz  —  10xsyz3  by  —xyz.  —8x2y2z2  +  10#4?/V. 

8.  abc  —  a2bc  —  ab2c  by  —abc.     —a2b2c2  +  a3b2c2  +  a2b3c'i. 

9.  x  +  7  by  x  —  10.  or  —  3#  —  70. 

10.  x  +  9  by  x  -  7.  x2  +  2x  -  63. 

11.  2x  -  3  by  x  +  8.  2a;2  +  13a;  -  24. 

12.  2a;  +  3  by  x  -  8.  2a;2  -  13x  -  24. 

13.  x3  —  7a;  +  5  by  a;2  —  2x  +  3. 

Ans.  x5  -  2a;4  -  4a;3  +  19a;2  -  31a;  +  15. 

14.  a2  -  5a6  -  b2  by  a2  +  hab  +  b2. 

Ans.  a4  -  25a262  -  10a&3  -  b\ 

15.  a;2  —  xy  +  x  +  y2  +  y  +  1  by  x  +  y  —  1. 

^.ws.  a;3  +  Sxy  -\-  ys  —  1. 

16.  a2  +  Zr  +  c2  —  6c  —  ca  —  a6  by  a  +  6  +  c. 

^ws.  a3  +  63  +  c3  —  3a6c. 


56  EXAMPLES. 

17.  3cr  +  2a  +  2a3  +  1  +  a4  by  a2  -  2a  +  1. 

^?is.  a6  —  2a3  +  1. 

18.  —  a.r2  +  Baasy3  —  day1  by  —ax  —  Say2. 

Ans.  a¥  +  27«y. 

19.  -2arty  +  ?/4  +  3ar#2  +  a4  -  2a;#3  by  a2  +  2«y  +  ?/2. 

^ws.  x6  +  2a;2?/3  +  y*. 

20.  2aa;+a>2+a2bya2+2aa;— a?2.  a4+4a3a;+4a2a;2-a;4. 

21.  262+3a6— a2by7a— 56.  -1068— a62+26a26— 7a8. 

22.  a2-ab+b2  by  a2+a6-62.  a4_a36a+2a68— 6*. 

23.  ±x2-3xy-y2by3x-2y.  12x»-17x2y+3xy2+2y3. 

24.  x5— xiy-\-xyi  —  ?/5  by  x+y.  xG—xiy2+x2yi—y6. 

25.  a;4+2afy-f4a;2?/2+8a;?/3+16?/4  bya— 2y.  x5-S2y5. 

26.  9ajy+27afy+81»*+3a;y8+y4  by  3a;-?/.       243a--5-?/5. 

27.  x+2y—Bzbyx-2y+Sz.  x*—4y2+12yz—9z*. 

Write   down   tbe   values    of    the    following   products   by 
inspection. 

28.  (a;  +  7)  (a?  +  1).  Ans.  x2  +  8a;  +  7. 

29.  (x  —  7)  (a?  +  14).  a;2  +  7a;  -  98. 

30.  (a  +  36)  (a  -  26).  a2  +  ab  -  66s. 

31.  (a  -  6)  (a  +  13).  a2  +  7a  -  78. 

32.  (a  +  2)  (a  -  12).  a2  -  10a  -  24. 

33.  (x  -  l)(x  +  12).  x2  +  11a;  -  12. 

34.  (2a;  -  5)  (a;  -  2).  2a;2  -  9a;  +  10. 

35.  (3a;  -  l)(a?  +  1).  3a;2  +  2a;  -  1. 

36.  (3a;  +  7)  (2a;  -  3).  6a;2  +  ox  -  21. 

37.  (3a;  +  8)  (3a;  -  7).  9a;2  +  3a;  -  56. 

38.  (2a;  +  7)  (5a;  +  4).  lOa?"  +  43a;  +  28. 

Solve  the  following  examples  by  formula1  (1),  (2),  (3)  in 
Art.  41. 

39.  (x2+xy-\-y2)  (x2— xy—  y2).      Ans.  xi—x2y2  —  2xy8—yi 

40.  (x2+xy-y2)(x2-xy  +  >r).  xi-x2y2+2xy*-yi 

4 1 .  (a;3 + 2x2 + 3a;  + 1 )  (a;3  -  2a'2 +  3a-  - 1 ) .    a-6  +  2a;4  +  5a;2 - 1 

42.  (a;-3)2(a;2+6a;+9).  a;4-18a;2+81 

43.  (x+y)2(x2— 2xy  —  y2).  a;4— 4a;2?/2  — 4a#3- 


EXAMPLES.  57 

44.  (2a?  +  Sy)2(ix2  +  12xy  -  9y2). 

Ans.  16s4  +  96xsy  +  144s2?/2  —  81y4. 

45.  (a»  +  by)  (ax  —  by)(a2x2  +  b2y2).  a¥  —  &V« 

46.  (as  +  6y)2(aa;  -  6?/)2.  oV  -  2a2b2x2y2  +  &V- 

Show  that  the  following  results  are  true  : 

47.  (as+62)  (c2+d2)  =  (ac+&d)2+  (ad-&c)2. 

48.  (a+&+c)2+a2+&2+c2=(a+&)2+(&+c)2+(c+a)2. 

49.  (a— 6)  (6— c)  (c— a)  =&c(c— 6)  +ca(a— c)+a&(&— a). 

50.  (a-&)3+&3-a3=3a&(&-a). 

51.  (a2+a&+&2)2-(a2-a&+&2)2=4a&(a2+&2). 

52.  (a-f&+c)3-a3-&3-c3=3(a+&)  (&+c)  (c+a). 

53.  (a+&)2+2(a2-&2)  +  (a-6)2=4a2. 

54.  (a-i)3+(&-c)3+(c-a)3=3(a-&)(6-c)(c-a). 

55.  (a-&)3+(a+o)3+  3(a  -  &)2(a+&)  +  3(a+6)2(a-6) 
=  (2a)3. 

56.  (a  +  &)2(&  +  c  -  a)  (c  +  «_&)  +  (a  -  &)2(a  +  &  +  c) 
(a+6— c)=4a6c2. 

57.  [(as  +  by)2  +  (ay  -  bxy]\_(ax  +  by)2  -  (ay  +  to)2] 
=  (a4-&4)(s4-?/4). 


58      THE   DIVISION    OF   ONE   MONOMIAL   BY  ANOTHER. 


CHAPTER    V. 

DIVISION. 

44.  Division  in  Algebra  is  the  process  of  finding,  from 
a  given  product  and  one  of  its  factors,  the  other  factor  ;  or 
it  is  the  process  of  finding  how  many  times  one  quantity  is 
contained  in  another. 

Division  is  therefore  the  converse  of  multiplication. 

The  Dividend  is  the  given  product ;  or  it  is  the  quantity 
to  be  divided. 

The  Divisor  is  the  given  factor ;  or  it  is  the  quantity  by 
which  we  divide. 

The  Quotient  is  the  required  factor;  or  it  is  the  number 
which  shows  how  many  times  the  divisor  is  contained  in  the 
dividend. 

The  above  definitions  may  be  briefly  written 

quotient  X  divisor  =  dividend, 

or  dividend  -=-  divisor  =  quotient. 

It  is  sometimes  better  to  express  this  last  result  as  a 

fraction ;  thus  dividend  ^.     , 

—-—, =  quotient. 

divisor 

It  is  convenient  to  make  three  cases  in  Division,  (1)  the 
division  of  one  monomial  by  another,  (2)  the  division  of  a 
polynomial  by  a  monomial,  (.'))  the  division  of  one  poly- 
nomial by  another. 

45.  The  Division  of  one  Monomial  by  Another. 
—  Since  the  product  of  4  and  z  is  4x,  it  follows  that  when 
4x  is  to  be  divided  by  .»■  the  quotient  is  i.     <  >r  otherwise 

4x  -7-  x  =  4. 
v\lso  since  the  product  of  a  and  b  is  ah.  the  quotient  of  ah 
divided  by  a  is  b  ;  that  is 

ab  -i-  a  =  b. 


THE  DIVISION   OF   ONE  MONOMIAL   BY  ANOTHER.      59 

Similarly  abc  -f-  a  =  be ;  abc  -r-  b  =  ac ;  abc  -+-  c  "=  a& ; 
abc  -r-  a&  =  c  ;  a&c  -=-  &c  =  a  ;  abc  -h  ca  =  b.  These  results 
may  also  be  written 


abc 
a 

= 

be  ; 

abc  _ 
6 

ac ; 

a&c  _ 
c 

rt& 

abc 
ab 

= 

c; 

a&c  _ 
be  ~ 

a ; 

a&c  _ 

ca 

&. 

Also  36a6 

9a4 

= 

36a6  _ 

36aaaaaa 

4a 

=  4aa,  by  remov- 
9a4  daaaa 

ing  from  the  divisor  and  dividend  the  factors  common  to 

both,  just  as  in  Arithmetic. 

Therefore  36a6  -=-  9a4  =  4a2. 

45aaaabbbcc 


Similarly       45a4&3c2  -=-  9a26c2  = 


9aa6cc 
5a2&2. 


Hence  we  have  the  following 
Rule. 

To  divide  one  monomial  by  another,  divide  the  coefficient  of 
the  dividend  by  that  of  the  divisor,  and  subtract  the  exponent 
of  any  letter  in  the  divisor  from  the  exponent  of  that  letter  in 
the  dividend. 

For  example    12xhf  -r-  \2xhf  =  6x5~3y3~2 
=  Gx^y. 

Also  Soa^Y  -=-  Ua2xif  =  5a2xy. 

Rem.  —  If  the  numerical  coefficient,  or  the  literal  part  of  the 
divisor  he  not  found  in  the  dividend,  we  can  only  indicate  the  division. 
Thus  if  la  is  to  be  divided  by  2c,  the  quotient  can  only  be  indicated 
by  la  +  2c  or  by  — .  In  some  cases,  however,  we  may  simplify  the 
expression  for  the  quotient  by  a  principle  already  used  in  Arithmetic. 
Thus  if  16a362  is  to  be  divided  by  12abc,  the  quotient  is  denoted  by 

16a8&2.     Here  the  dividend  =  4a&  x  4r/2&  and  the  divisor  =  4ab  x  3c; 
12abc 

thus  the  factor  4o6,  which  occurs  in  both  dividend  and  divisor,  may 
be  removed  in  the  same  way  as  in  Arithmetic,  and  the  quotient  will  be 

denoted  by  — .     That  is 
3c 

Wash2  _  4ab  x  4rt2ft  _  4a~b 
V2abc  ~     4ab  x  3c  3c  ' 

by  removing  the  common  factor  4o6. 


60  THE  RULE   OF  SIGNS — EXAMPLES. 

NoTBf.  —  If  we  apply  the  above  rule  to  divide  any  power  of  a  letter 
by  the  same  power  of  the  letter,  we  are  led  to  a  curious  conclusion. 
Thus  by  the  rule 

a3  -f  a3  =  a8~3  =  a0; 

but  also  o3  -f  a3  =z  —      =1, 

a3 

by  removing  the  common  factor  a3; 

.  •.  cfi  =  1 ; 
that  is,  any  quantity  whose  exponent  is  0  is  equal  to  1. 

The  true  significance  of  this  result  will  be  explained  in 
Art.  119. 

46.   The  Rule  of   Signs  for  division  may  be  obtained 
from  an  examination  of  the  cases  which  occur  in  multiplica- 
tion, since  the  product  of  the  divisor  and  quotient  must  be 
equal  to  the  dividend. 
Thus  we  have 

+a  x  (  +  &)  =  +ab, 
—a  x  (  +  6)  =  — aft, 
+a  X  (  —  b)  =  —  ab, 
—a  x   (-&)  =  +ab, 

Hence  in  division  as  well  as  in  multiplication,  like  signs 
produce  +,  and  unlike  signs  produce  — . 

EXAMPLES. 
Divide 

1.  Sa%  by  4a&.  Ans.  2a. 

2.  —  15xy  by  Sx.  —5?/. 

3.  -21«263by  -la%\  36. 

4.  45a6o2.«4  by  -9a3bx2.  —5a*bx*. 

5.  -36aWby  — 24aW.  $a&9. 

6.  3a3bya2.  "  Sx. 

7.  -58aWby  -2<r7,.  29aW. 

8.  63a4a;y  by  -  7a V-  -9a6. 

9.  1. <■'"+"  by  2aJ".  2#". 
10.   -77a",+2by  lis"*.  -7.v,-. 


+a&  -*- 

+&  =  -fa. 

-ab  -*■ 

+  />  =  —a. 

-aft  - 

-b  =  +a. 

+a&  h- 

—  b=  —a. 

TO   DIVIDE   A    POLYNOMIAL   BY  A    MONOMIAL.  61 

47.  To  Divide  a  Polynomial  by  a  Monomial. 

Since  (a  —  b)c  =  ac i  —  be  ; 

,,       a                           ac  —  be  7 

therefore  =  a  —  o. 

c 

Also  since    (a  —  fr)   x   (— c)  =  —ac  +  6c; 

therefore  ~ac  +  6c  =  a  -  b. 

—  c 

Also  since  (a  —  6  +  c)a&  =  a25  —  ab2  +  a&c ; 

,,       fi  a?b  —  ab2  +  abc  ,    , 

therefore      — =  a  —  b  +  c. 

a& 

Hence  we  have  the  following  rule : 

To  divide  a  polynomial  by  a  monomial,  divide^  each  term  of 

the  dividend  sejmrately  by  the  divisor. 

For  example 

(8a3  -  Ga2b  +  2a2c)  -=-  2a2  =  4a  -  3b  +  c. 

(9a;  -  12y  +  3z)  -s-  —  3  =  -3a:  +  4y  -  z. 

(36a362  -  24a265  -  20a4Z>2)  -j-  4a2&  =  9ab  -  6b4  -  5a2b. 

(2x2  —  5xy  —  \x2yz)  -= \x  =  —Ax  +  10?/  +  3a;?/3. 

EXAMPLES. 

Divide 

1.  —  35a;6  by  7a3.  Ans.  —  5a8. 

2.  a3?/3  by  arfy.  a;?/2. 

3.  4aW  by  a&V.  4ac. 

4.  a2  —  2xy  by  .r.  a;  —  2y. 

5.  a;6  -  7a;5  +  4a4  by  a;2.  x*  -  7x3  -f-  4a;2. 

6.  10a;7  -  8a;6  +  3a;4  by  a;3.  10a;4  -  8a;3  +  3a;. 

7.  15a;5  -  25a;4  by  -  5a;3.  -3a;2  +  ox. 

8.  27a;6  -  36a;5  by  9a;5.  3a;  -  4. 

9.  -24a;6  -  32a;4  by  -8a;3.  3a;3  +  4a;. 
10.  a2  —  ab  —  ac  by  —a.  —a  +  b  +  c. 

48.  To  Divide  one  Polynomial  by  Another.  —  Let 

it  be  required  to  divide  a3  +  2a2  —  3a  by  a2  +  3a. 

Here  we  are  to  find  a  quantity  which  when  multiplied  by 
the  divisor  will  produce  the  dividend.     Hence  the  dividend 


62  TO   DIVIDE   ONE  POLYNOMIAL   BY  ANOTHER. 

is  composed  of  all  the  partial  products  arising  from  the 
multiplication  of  the  divisor  by  each  term  of  the  quotient 
(Art.  39).  Arranging  both  the  dividend  and  divisor  accord- 
ing to  descending  powers  of  a,  we  see  that  the  first  term  a3 
of  the  dividend  is  the  product  of  the  first  term  a2  of  the 
divisor  by  the  first  term  of  the  quotient  (Art.  43) ;  therefore, 
dividing  a3  by  a'2  we  obtain  a  for  the  first  term  of  the 
quotient.  Multiplying  the  whole  divisor  by  a  we  obtain 
ft3  _j_  3a2  for  the  partial  product  of  the  divisor  by  the  first 
term  of  the  quotient ;  subtracting  this  product  from  the 
dividend  we  obtain  the  first  remainder  —a2  —  3a,  which  is 
the  product  of  the  divisor  by  the  remaining  terms  of  the 
quotient,  and  consequently  the  first  term  —a2  of  this  product 
is  the  product  of  the  first  term  of  the  divisor  by  the  second 
term  of  the  quotient.  Dividing  therefore  this  first  term  —a2 
by  the  first  term  of  the  divisor  a2  we  obtain  —1  for  the 
second  term  of  the  quotient.  Multiplying  the  whole  divisor 
by  —1  we  obtain  —a2  —  3a  for  the  product  of  the  divisor  by 
the  second  term  of  the  quotient ;  subtracting  this  product 
there  is  no  remainder.  As  all  the  terms  in  the  dividend 
have  been  brought  down,  the  operation  is  completed.  Hence 
a  —  1  is  the  exact  quotient. 

The  work  may  be  arranged  as  follows  : 

Divisor.  Dividend.  Quotient. 

a2  +  3a)  a3  +  2a2  -  3a  (a  -  1 

a3  +  3a2 

-  a2  -  3a 

-  a2  -  3a 


It  will  be  observed  that  in  getting  each  term  of  the 
quotient  in  this  example,  we  divide  that  term  of  the  divi- 
dend containing  the  highest  power  of  a  by  the  term  of  the 
divisor  containing  the  highest  power  of  the  same  letter , 
and  therefore,  when  the  dividend  and  divisor  are  arranged 
according  to  descending  powers  of  a,  any  term  of  the 
quotient  is  found  by  dividing  the  first  term  of  the  divisor 


TO   DIVIDE    ONE  POLYNOMIAL    BY  ANOTHER.  G3 

into  the  first  term  of  the  dividend,  or  into  the  first  term  of 
one  of  the  remainders. 

Hence  for  the  division  of  one  polynomial  by  another,  we 
have  the  following 

Rule. 

Arrange  both  dividend  and  divisor  according  to  ascending 
or  descending  powers  of  some  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor,  and  write  the  result  for  the  first  term  of  the 
quotient;  multi-ply  the  whole  divisor  by  this  term,  subtract 
the  product  from  the  dividend,  and  to  the  remainder  join  as 
many  terms  from  the  dividend,  taken  in  order,  as  are  required. 

Divide  the  first  term  of  the  remainder  by  the  first  term  of 
the  divisor,  and  write  the  result  for  the  second  term  of  the 
quotient ;  multiply  the  whole  divisor  by  this  term,  and  subtract 
the  product  from  the  last  remainder. 

Continue  this  operation  until  the  remainder  becomes  zero, 
or  until  the  first  term  of  the  remainder  will  not  contain  the 
first  term  of  the  divisor. 

This  method  of  dividing  is  similar  to  long  division  in 
Arithmetic,  i.e.,  we  break  up  the  dividend  into  parts,  and 
find  how  often  the  divisor  is  contained  in  each  part ;  and  then 
the  sum  of  these  partial  quotients  is  the  complete  quotient. 
Thus,  in  the  example  just  solved,  a3  +  2a2  —  3a  is  divided 
by  the  above  process  into  two  parts,  viz.,  a3  +  3a2,  and 
—  a2  —  3a,  and  each  of  these  is  divided  by  a2  +  3a,  giving 
for  the  partial  quotients  a  and  —1  ;  thus  we  obtain  the 
complete  quotient  a  —  1. 

Note.  —  The  divisor  is  often  put  on  the  right  of  the  dividend  and 
the  quotient  beneath  the  divisor  as  follows : 
Dividend.  Divisor. 

a2  +  2ab  +  b2  \a  +  b 
a2  +    ab  a  +  b  Quotient. 

ab  -f-  b2 
ab  +  b2 


64  EXAMPLES. 

It  is  of  great  importance  to  arrange  both  dividend  and 
divisor  according  to  ascending  or  descending  powers  of  some 
common  letter ;  and  to  attend  to  this  order  in  every  part  of  the 
operation. 

EXAMPLES. 

1.  Divide  24a*2  -  6oxy  +  2\y2  by  8a;  -  By. 
The  operation  is  conveniently* arranged  as  follows  : 

8x  —  3?/)  24a*2  —  Goxy  +  21y2(3x  —  ly. 
24a*2  -    9xy 

—  56a:?/  +  2\y2 

—  56a*?/  +  21y2 

Divide 

2.  x2  +  Sx  +  2  by  x  +  1.  Ans.  x  +  2. 

3.  x2  —  7x  +  12  by  x  —  3.  x  —  4. 

4.  x2  —  \\x  -f  30  by  x  —  5.  a  —  6. 

5.  a;2  -  49a:  +  600  by  x  -  25.  as  -  24. 

6.  3a*2  +  10a*  +  3  by  x  +  3.  3a*  +  1. 

7.  2a*2  +  11a*  +  5  by  2x  +  1.  a?  +  5. 

8.  5a*2  +  Ha;  +  2  by  a*  +  2.  5a?  +  1. 

9.  2a*2  +  17a*  +  21  by  2a*  +  3.  a>  +  7. 

10.  5a*2  +  16a:  +  3  by  a:  +  3.  5a;  +  1. 

11.  Divide  3a4  -  10a86  +  22a262  -  22ab3  +  15&4  by 
a2  _  2ab  +  3b2. 

The  operation  is  written  as  follows. 
aa  _  2ab  +  36a)3a4  -  10a8ft  +  22a2&2  —  22a&3  +  156*(3a9  -  iab  +  5b- 
3a4-    6a86  +    9a2&2 

-  4a3i  +  13a2&2  -  22r//,3 

-  4o*6  +    8«-V-a  -  12a68 

5a262  -  10a&8  +  15&4 
50s%a_  I0a68+  15/>4 

12.  Divide  a*7  -  5a*5  +  7a*8 + 2a:2  -  6a:  -  2  by  1  +  2a*  -  3.x*2  +  & . 
Arrange  both  dividend  and  divisor  according  to  descending 

powers  of  x,  and  arrange  the  work  as  follows  : 


X' 


EXAMPLES.  65 

-  5a;5  +  1.x?  +  2x2  -  6x  -  2  la;4  -  3a;2  +  2x  +  1 


a;7  —  3a5  4-  2.c4  +    «3  »3  —  2a; 


-  2a;5- 

-  2X5 

-  2x* 

+  Gx3 

+  Gx3 

+  2x- 

-    l.r- 

-  Gx 

-  2x 

-  2x* 

-  2a;4 

+  Gx2 

-  Ax  - 

-  Ax  - 

-  2 

-  2 

"We  might  have  arranged  the  dividend  and  divisor  accord- 
ing to  ascending  powers  of  x  as  follows  : 
_2  -  6a>  +  2a;2  +  7a;3  -  5a;5  +  a;7 1 1  -f-  2a;  -  3a;2  +  a;4 

_2  -  Ax  +  6a;2  -  2a;4  -2  -  2a;  4-  a;3 


-  2a; 

-  2x 

-  Ax2  +  Ix3  +  2a;4  -  5a;5 

-  4a;2  +  6a;3             -  2a;5 

x3  +  2a;4  -  3a;5  +  x1 
x3  +  2a;4  -  3a;5  +  x? 

We  thus  obtain  the  same  quotient  we  had  before,  though 
the  terms  are  in  a  different  order. 
13.   Divide  a3  +  b3  +  c3  -  Sabc  by  a  +  b  +  c. 
Arrange  the  dividend  according  to  descending  powers  of  a. 
Sabc  4-    63  +  c8[a    +    & 


a3  +     a2b  +  «2c 

a2  —  ab  —  ac  +  &2  —  be  +  c2 

—  a26  —  a2c  — 

—  a2b  —  a&2  — 

3a6c 

abc 

-  o2c  + 

—  a?c 

ab2  —  2a,bc 

—    abc  —  ac2 

air  —     abc  +  ac2  +  ft3 

o62                          +  63    4-  &*c 

—  a&c  4-  8C2  —  ?;2c 

—  a&c              —  62c  —  6c2 

ac2  4-  6c2  4-    c3 
ac2  +  &C2  4-    c« 

66  EXAMPLES. 

Iii  this  example  we  arrange  the  terms  according  to  de- 
scending powers  of  a  ;  then  when  there  are  two  terms,  such 
:is  d-b  and  a?c,  which  involve  the  same  power  of  a,  we 
select  a  new  letter,  as  6,  and  put  the  term  which  contains  b 
before  the  term  which  does  not ;  and  again  of  the  terms  air 
and  a&c,  we  put  the  former  first  as  involving  the  higher 
power  of  b. 

14.    Divide  x4  +  4a4  by  ar  +  2ax  +  2cr. 

x4  +  4«4      .  |  a2  +  2ax  +  2a2 


X* 

+ 

2ax3 

2a.x3 
2ax* 

+  2a2x2 

-  2«2x2 

—  4a2x2 

x2  —  2ax  -f- 
—  4a3x 

2a2 

.2a2x2 
2a2x2 

+  4a3x  +  4a4 
+  4a3x  +  4a4 

Divide 

15.  x5  -  5.x4  +  9x3  -  6x2  -  a  +  2  by  a2  -  Sx  +  2. 

^dris.  x3  —  2a;2  +  x  +  1. 

16.  x5  —  2x4  -  4.x3  +  19»j2  -  31a  +  15  by  x3  -  7x  +  5. 

^4?)s.  a;2  —  2x  +  3. 

17.  u3  +  &3  +  3a&c  —  c3  by  a  +  &  —  c. 

Ans.  <r  —  ab  +  ac  +  6'2  +  be  +  c2. 

18.  xl  +  64  by  x2  +  4.x  +  8.  x2  -  4x  +  8. 

19.  a6  -  66  by  a3  -  2a2b  +  2a&3  -  &3. 

.4ns.  a8  +  2aa6  +  2a6a  +  b\ 
When  the  coefficients  are  fractional  we  may  still  use  the 
ordinary  process,  combining  the   fractional   coefficients  by 
the  rules  of  Arithmetic. 

20.  Divide  ix3  +  r^xy2  +  &f  by  hx  +  £</. 

$x*  +  TVx?/2  +  -&y*  \\x  +  \y 

K  +  Ky  K  -  ^  +  fcf 

—  t>-';7/  +  tV'.V2 

•,7r  +  A»" 

W2  +  fay* 


DIVISION    WITH   THE   AID    OF  PARENTHESES.  67 

Divide 

21.  %a3  -  fa%s  +  2£aa?  -  27a*8  by  \a  -  Sx. 

Arts.  \<&  -  Sax  +  9a2. 

22.  fra*  -  TVr  +  Ty<  -  A  b>T  ia  ~  i-     K  ~  ia  +  tV 

49.  Division  with  the  Aid  of  Parentheses.  —  Some- 
times it  is  found  convenieut  to  divide  with  the  aid  of  paren- 
theses thus  : 

1.  Divide  x3—  (a+b-\-c)x2+  (ab-\-ac+bc)x—abc  by  x—c. 
x3  —  (a  +  b  +  c)ar+  (ab  +  ac  -f-  bc)x  —  abc  \x  —  c 

x3  —  ex2  a*2—  (a+b)x+ab 

—  (a  +  b)x2  +  (ab  +  ac  +  be)  x 

—  (a  +  b)x2  +  (a  +  6)  ex* 

abx  —  abc 
abx  —  abc 
Divide 

2.  a2a*4+(2ac— 62)a*2+c2  by  ax2— bx+c.    Ans.  ax2+bx+c. 

3.  x3  +  (a  +  &  +  c)a:2  +  (a&  4-  ac  +  6c)  a  +  abc  by  a  +  &• 

-4ws.  a2  +  (a  +  c)a;  -+-  ac. 

4.  ax3  —  (a2  +  &)ar  +  b2  by  ckc  —  b.  x2  —  aa*  —  &. 

5.  a2(6  +  c)  +  62(a  —  c)  +  c2(a  —  6)  +  a&c  by  a  +  &  +  c. 

.4ns.  a  (6  +  c)  —  6c. 

50.  "Where  the  Division  cannot  be  Exactly  Per- 
formed. —  In  the  examples  given  thus  far  the  divisor  has 
been  exactly  contained  in  the  dividend.  It  may  happen,  as 
in  Arithmetic,  that  the  division  cannot  be  exactly  performed. 
In  such  cases  it  is  a  good  exercise  for  the  student  to  deter- 
mine the  accurate  value  of  the  remainder. 

Divide  a3  -  6a*2  +  lia*  +  2  by  a*  —  2. 
x3  —  6a:2  +  11a*  +  2  [as   —  2 
a;3  -  2x2  x2  -  4a*  +  3 


—  4a*2  + 

-  4a*2  -f- 

llaj 

8a* 

3a*  +  2 
3a;  -  6 

68 


IMPORTANT   EXAMPLES  IN  DIVISION. 


The  division  can  be  carried  no  further  without  fractions, 
because  x  will  not  go  into  8.  We  therefore  express  the 
result  in  the  same  way  as  in  Arithmetic,  that  is,  by  adding 
to  the  quotient  a  fraction  of  which  the  numerator  is  the 
remainder  and  the  denominator  the  divisor.  Thus  the  result 
is 

*3-<^  +  ll*  +  2  =  a*  -  Ax  +  3  +  1A_. 
x  —  2  x  —  2 


EXAMPLES. 


Find  the  remainder  when 


x3  -  6a;2  +  12a;  -  17  is  divided  by  x  -  3.    .4ns. 
3a;3  —  7a;  —  9  is  divided  by  x  +  1. 

5a;2  —  4a;  —  7  is  divided  by  a;  +  2. 

7ar  —  3a;  —  33  is  divided  b}'  4x  —  5. 
27a;3  -f  9ar  —  3a;  —  5  is  divided  by  3x  —  2. 
1 6a;3  -  19  +  39a;  -  46a;2  is  divided  by  8x  -  3. 
7.    8a;  —  8a;2  +  5a;3  +  7  is  divided  by  5a;  —  3. 


2a;8  + 
4a;3 


-8. 
—  5. 

5. 
-18. 

5. 

-10. 

10. 


51.  Important  Examples  in  Division.  —  The  follow- 
ing examples  are  very  important ;  they  may  be  easily  verified, 
and  should  be  carefully  noticed. 

=  x    +  ?/, 

=  x2  +  xy   +  y2, 

=  x3  +  x'hj  +  xy2  +  y», 
x  —  y 

and  so  on ;  the  terms  in  the  quotient  all  being  jiositive. 

-x2  -  y2 


.1- 

— 

:'/- 

X 

- 

y 

Xs 

— 

y3 

X 

- 

y 

X* 

— 

t 

11. 


x3  —  cry  +  x>/'2 


x  +  y 
x*  —  y4 
x  +  y 
~  -L  =  .t-5  —  a;4^  -f  x8y2  —  an/8  -j-  xy*  —  ?/6, 


.(•:i 

+  ?/3 

X 

X6 

+  y 
+  yh 

X 

x> 

+  y 

IMPORTANT  EXAMPLES   IN  DIVISION.  69 

and  so  on  ;    the  terms  in  the  quotient  being  alternately  posi- 
tive and  negative. 

3 

1  -  xy  +  y\ 
:  a4  -  xhj  +  xY  -  vy*  +  y\ 
>-x5y  +  x4y2  -  xY  4-  A4  -  xy5  +  y\ 

x  +  y 

and   so   on ;    the   terms   in   the   quotient    being   alternately 
positive  and  negative. 

The  student  can  verify  these  results  in  any  particular  case, 
and  carry  on  these  operations  as  far  as  he  pleases,  and  he 
will  thus  gain  confidence  in  the  truth  of  the  following  state- 
ments for  which  we  shall  hereafter  give  a  general  proof.  See 
Art.  170. 

These  different  cases  may  be  conveniently  arranged  in  the 
following  concise  statements  : 

xn  —  yn  is  divisible  by  x  —  y  if  n  be  any  whole  number. 
(1)  That  is :  the  difference  of  any  two  equal  poivers  of  two 
numbers  is  ahvays  divisible  by  the  difference  of  the  two 
tumbers. 

xn  —  yn  is  divisible  by  x  +  y  if  n  be  any  even  whole 
number.  (2)  That  is  :  the  difference  of  any  tivo  equal  even 
powers  of  tivo  numbers  is  always  divisible  by  the  sum  of  the 
numbers. 

xn  +  yn  is  divisible  by  x  -\-  y  if  n  be  any  odd  whole 
number.  (3)  That  is  :  the  sum  of  any  two  equal  odd  powers 
of  two  numbers  is  always  divisible  by  the  sum  of  the  numbers. 

xn  -\-  yn  is  never  divisible  by  x  +  y  or  x  —  y,  when  n  is 
an  even  whole  number. 

EXAM  PLES. 

Write  the  results  in  the  following  by  these  three  state 
ments. 

1.  a5  —  b5  h-  a  -  6.     Ans.  a*  +  azb  +  a%-  +  ab*  +  b*. 

2.  X3   -    1    -r-  X  -    1.  JB2  +   X   +    1. 


TO  EXAMPLES. 

3.  ga*  -  bz  -H  2a  -  h.  Ans   4a2  +  2a&  +  b\ 

4.  8a;3  -  27?/8  -*•  2x  -  :'.?/.  4a;2  +  6xy  +  9?/2. 

5.  a;4  —  16?/4  -s-  a?  -f  2y.  a;3  -  2a;2?/  +  4a;?/2  —  8?/3. 

6.  16a;4  —  y4  -j-  2x  +  y.  8a;8  —  4a;2?/  +  2xy~  —  yz. 

7.  a;3  +  1  -=-  x  +  1.  a;2  -  a?  +  1. 

8.  a3  +  G4  -s-  a  +  4.  a2  -  4a  +  16; 

9.  8a;3  +  27 y3  -=-  2a;  +  Sy.  4a;2  —  6a;?/  +  9y-. 
Divide 

10.  3a;3  -  9a;2?/  -  \2xif  by  -3a;.  -a;2  +  Sxy  +  4?/2. 

1 1 .  4a;4?/4  —  8a;3?/2  +  6a;?/3  by  -  2a;?/.      —  2a;3?/3  +  4a;2?/  —  Sy2. 

12.  xsy  —  3a; V  4-  4a-?/3  by  a;?/.  a;2  —  3xy  -f  4?/2. 

13.  -15a868  -  3a262  +  12a6  by  -3o6.       5a262  +  a6  -  4. 

14.  60a  W  -  48a  W  +  36aW  by  4a6c2. 

Ans.  15a262  -  12a&8  +  9«6c2. 

15.  _3a2  +  %ab  -  6ac  by  -fa.  2a  -  36  +  4c. 
1G.  fa;5?/2  -  3a-3?/4  -  Go-4?/8  by  -fa;3?/2.      -3a-2  +  2y2  +  4a;?/. 

17.  £a2a;  —  -^abx  —  facaJ  by  faa;.  |a  —  £6  —  c. 

18.  a;2  -  11a;  +  30  by  x  -  5.  x  -  6. 

19.  x-  -  7a-  +  12  by  x  -  3.  a-  -  4. 

20.  3a;2  +  a-  -  14  by  a:  -  2.  3a;  +  7. 

21.  6a;2  -  31a;  +  35  by  2a;  -  7.  3a;  -  5. 

22.  15a2  +  17aa;  -  4ar  by  3a  +  Ax.  ha  —  x. 

23.  60a;2  -  4a;?/  -  45?/2  by  10a;  -  9y.  6x  +  by. 

24.  -4xy  -  15y2  +  96a;2  by  12a-  -  by.  8a;  +  By. 

25.  100a;3  -  3a-  -  13a-2  by  3  +  25a;.  4a-2  -  x. 

26.  7a;8  +  96a-2  -  28a-  by  7x  -  2.  a-2  +  14.r. 

27.  a;5  -  4a;4  +  3x3  +  3ar  -  3a;  +  2  by  a;2  -  a-  -  2. 

Ans.  Xs  —  3a-2  4-  2a-  —  1. 

28.  x6+x4y—xsy2-{-xs—2xy2+yshyx2+xy—y-.     xs+x—y. 

29 .  2a;8  -  8a-  +  a;4  +  12-  7ar  by  x2+  2  -  3a-.      ar+  5a;  +  6 . 

30.  8a;8  -  4a;2-  \2*x  +  a;4  -  192  by  a-2-  16.     a-2+  8x  +  12. 

3 1 .  a;9  —  if  by  a;2  +  xy  4  y2.     a;7  —  x6y  +  x*y*  —  x*y* + xy*  —  if' . 

32.  2a4+27a&8— 816*  bya+36.     2a8-6a2&+18a&9— 2768. 

33 .  a;5 +  x*y  4  xhf  4  xhf  +  xy* + y6  by  b8 + y8.       a-2 + ay/  +  >/-. 

34.  u5  +  2  A  +  Safy2  -  a;2?/8  -  2xy*  -  Sy"  by  r1  -  ?/8. 

An 8.  x-  4-  2ajf  +  8y". 


EXAMPLES.  71 

35.  x12  +  x*  -  2  by  a;4  +  a;2  +  1.     .dn.s.  x*  -  x6  +  2a;2  -  2. 

36.  a;4  —  x3y  —  xyz  +  y4  by  a;2  +  xy  +  y2.      a?2  —  2#y  +  y2. 

37.  49a;2+ 21a;?/+ 12?/z  -  16z2  by  7x  +  3y—4z.        7x  +  4z. 

38.  a2  +  2a&  +  62  -  c2  by  a  +  6  -  c.  a  +  6  +  c. 

39.  a3  +  3a26  +  &3  -  1  +  Bab2  by  a  +  b  -  1. 

J.ws.  a2  +  2a&  +  6*  +  a  +  6  +  1. 

40.  a;8  —  ys  by  a--3  +  x2y  +  xy2  -f  y3.      x5  —  x4y  +  xy*  —  y5. 

41.  a12  +  2a666  +  bu  by  a4  +  2a2b2  +  i4. 

^ws.  a8  -  2«662  +  3a464  -  2a2&6  +  &8. 

42.  §a2c8  +  Tf  ^a5  by  £a2  +  \ac.        fta*  -  fa2c  +  fac2. 

43.  T^a4_|a3-|a2+|a+^byfa2-|-a.      ^_ia-§. 

44.  36a;2  +  ±y2  +  \  -  4xy  -  6x  +  %y  by  6»  -  \y  -  f 

.4ns.  6a;  —  -|y  —  £. 

45.  (a;  +  ?/)2— 2(z  +  ?/)z +  22bya;  +  ?/  — z.       ar  +  y  —  z. 

46.  aa;2  —  a&2  -f-  b2x  —  a;3  by  (x  +  6) (a  —  a;).       a;  —  6. 

47.  (b-c)a3+(c-a)bz+(a-b)cs  by  a2—ab—ac+bc. 

Ans.  a{b-c)  +  b2-c2. 

48.  a;6—  1  by  a;  -  1.     (Art.  51.)  a;G+  a,4+a;3+a;2-|-a;+l. 

49.  a;4  —  81y4  by  x  —  By.  x3  +  3a;2y  +  9xy2  +  27y3. 

50.  a;5  —  ?/5  by  a;  —  y.  a;4  +  x3y  -+-  a;2?/2  +  xy3  -f-  y*. 

51.  a9  -  69  by  a3  -  Z>3.  a6  +  a3b3  +  66. 

52.  27a;3  +  8?/3  by  3a;  +  2y.  9x2  -  6xy  +  4?/2. 

53.  64a;6  —  y6  by  2a;  —  y. 

Ans.  32a;5  +  16x*y  +  8afy2  +  ix2y*  +  2a;?/4  +  y5. 

54.  32a;5 +  y5  by  2a;  +  y.        16a;4-  8x-3y  +  4afya  —  2a;?/3 +  ?/4. 

55.  Divide  the  product  of  a2  +  ax  -f-  a;2  and  a3  +  a;8  by 
a4  +  a2x2  +  a;4.  J.«s.  a  +  a\ 

56.  Divide  (ax  +  %)2  +  (ay  -  &a;)2  +  c2a;2  +  c2y2  by 
a;2  +  y2.  Ans.  a2  +  b2  +  c2. 

57.  Divide  a?b  —  bx2+  a?x  —  x8  by  (a;+&)  (a— a;) .       a+x. 

58.  Divide  &(a;8  +  a3)  +  ax{x2  -  a2)  +  a8 (a;  +  a)  by 
(a  +  6)  (a;  +  a) .  ./1ms.  a;2  —  aa;  +  a2. 

59.  Divide  (x  +  y)7  -  (a7  +  y7)  by  (a;2  +  «y  +  y2)2. 

.4ms.  7xy{x  4-  y). 


72     EQUATION   OF   CONDITION —  UNKNOWN    QUANTITY. 


CHAPTER    VI. 

SIMPLE     EQUATIONS     OF    ONE     UNKNOWN 
QUANTITY. 

52.  Equations  —  Identical  Equations.  —  An  Equation 
is  a  statement  in  Algebraic  language  that  two  expressions 
are  equal.     Thus,        2a;  +  4  =  a?  +  8 

is  an  equation  ;  it  states  that  the  expression  2x  -f  4  is  equal 
to  the  expression  x  +  8. 

The  two  equal  expressions  thus  connected  are  called  sides 
or  members  of  the  equation.  The  expression  to  the  left 
of  the  sign  of  equality  is  called  the  first  side  or  member, 
and  the  expression  to  the  right  is  called  the  second  side  or 
member.     Every  equation  has  two  members. 

An  Identical  Equation,  or  briefly  an  Identity,  is  one  in 
which  the  two  members    are   equal  whatever  numbers   the 
letters  represent.     Thus,  the  following  are  identical  equa- 
tions :  „  '      ,    _ 
x  +  3  +  x  +  4  =  2x  -f-  7, 

(a  -f-  x)  (a  —  x)  —  a2  +  xz  —  0  ; 
that  is,  these  Algebraic  statements  are  necessarily  true, 
whatever  values  we  assign  to  x  and  a.  All  the  equations 
used  in  the  previous  chapters  to  express  the  relations  of 
Algebraic  quantities  are  identical  equations,  because  they 
are  true  for  all  values  of  these  quantities. 

53.  Equation  of  Condition  —  Unknown  Quantity. 
—  An  Equation  of  Condition  is  one.  which  is  true  only  when 
the  letters  represent  some  particular  value.  For  example, 
the  equation,  „  _  .., 

cannot  be  true  unless  x  —  5,  and  is  therefore  an  equation  of 
condition. 


AXIOMS.  73 

An  equation  of  condition  is  called  briefly  an  equation. 

The  letter  whose  value,  or  values,  it  is  required  to  find 
is  called  the  unknown  quantity.  Thus  x  is  the  unknown 
quantity  in  the  above  equation. 

To  solve  an  equation  means  to  find  the  value,  or  values, 
of  the  unknown  quantity  for  which  the  equation  is  true. 
These  values  of  the  uukuown  quantity  are  said  to  satisfy  the 
equation,  and  are  called  the  roots  of  the  equation. 

An  equation  which  contains  ouly  one  unknown  quantity  is 
called  a  simple  equation,  or  an  equation  of  the  first  degree, 
when  the  unknown  quantity  occurs  only  in  the  first  power. 
It  is  usual  to  denote  the  unknown  quantity  b}T  the  letter  x. 
The  equation  is  said  to  be  of  the  second  degree  or  a  quadratic 
equation  when  x2  is  the  highest  power  of  x  which  occurs,  and 
so  on.* 

Thus  2x  +  G  =  a;  +  8, 

and  ax  -\-  b  —  c 

are  simple  equatious. 

x2  -  2x  =  3 
is  a  quadratic  equation. 

54.  Axioms.  —  An  Axiom  is  a  self-evident  truth.  The 
operations  employed  in  solving  equations  are  founded  upon 
the  following  axioms  : 

1.  If  equal  quantities  be  added  to  equal  quantities,  the 
sums  will  be  equal. 

2.  If  equal  quantities  be  taken  from  equal  quantities,  the 
remainders  will  be  equal. 

3.  If  equal  quantities  be  multiplied  by  equal  quantities, 
the  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  equal  quantities,  the 
quotients  will  be  equal. f 


*  The  equation  is  supposed  to  be  reduced  to  such  a  form  that  the  unknown 
quantity  is  found  only  in  the  numerators  of  the  terms,  and  that  the  exponents  of  its 
powers  are  expressed  by  positive  integers. 

t  If  the  divisors  are  different  from  zero. 


74  CLEARING   OF  FRACTIONS. 

5.  Like  powers  and  like  roots  of  equal  quantities  are 
equal. 

These  axioms  ma}r  be  summed  up  in  the  following  one : 
If  the  same  operations  be  performed  on  equal  quantities,  the 
results  will  be  equal. 

In  the  solution  of  equations  there  are  two  operations 
of  frequent  use.  These  are  (1)  clearing  the  equation  of 
fractions,  and  (2)  transposing  the  terms  from  one  member 
to  the  other  so  that  the  unknown  quantity  shall  finally  stand 
alone  as  one  member  of  the  equation. 

55.  Clearing  of  Fractions.  —  Consider  the  equation 

X        X        x 

2^36 
Multiplying  each  term  by  2  X  3  X  6  (Axiom  3),  we  get 
3  X  6x  +  2  X     6s  +       2  X  3x*  =  2   X  3  x   G  X   2, 
or  18a-  +  12a:  +  6as  =  72; 

dividing  each  term  by  6  (Axiom  4),  we  get 

3a-  4-  2x  +  x  =  12, 
or  Gx  =12. 

Instead  of  multiplying  each  term  by  2  x  3  x  G,  we  might 
multiply  each  term  by  the   least  common    multiple   of   the 
denominators,  which  is  6,  and  get  immediately 
3x  +  2a;  +  x  =  12. 
Hence   to   clear  an   equation   of   fractions,   we    have   the 
following 

Rule. 

Multiply  each  term  of  the  equation  by  (he  least  common 
nulla  pie  of  the  denominators. 

Clear  the  following  equation  of  fractions : 

X    _|_    *    _    X     _     g 

3        5        7 
Here  the  least  common  multiple  of   the  denominators  is 
the  product  of  the  denominators,  3,  5,  and  7.     Multiplying 
each  term  by  it,  we  get 

35a;  +  21as  -  15a:  =  315, 
or  41a;  =  315. 


TRANSPOSITION.  75 

Clear  the  following  equation  of  fractions  : 
2  +  2  +  2  +  ±  =  4. 

4        6        8        12 
Here  24  is  the  least  common  multiple  of  the  denominators. 
Multiplying  each  term  by  it,  we  get 

6x  +  4a>  +  3a;  +  2x  =  96, 
or  15a?  =  96. 

56.  Transposition.  —  To  transpose  a  term  is  to  change 
it  from  one  member  of   an  equation  to  the   other  without 
destroying  the  equality  of  the  members. 
Suppose,  for  example,  that  x  —  a  =  b. 
Add  a  to  each  member  (Axiom  1)  ;  then  we  have 
x  —  a  -\-  a  =  b  +  a; 
therefore,  since  —  a  and  +a  cancel  each  other,  we  have 
x  =  b  +  a. 
Again,  suppose  that  x  +  b  =  a. 

Subtract  b  from  each  member  (Axiom  2)  ;  then  we  have 
x  +  b  —  b  =  a  —  b ; 
therefore,  since  +6  and  —b  cancel  each  other,  we  have 
x  =  a  —  b. 
Here  we  see,  in  these  two  examples,  that  —a  has  been 
removed  from  one  member  of   the  equation,   and   appears 
as  +a  in  the  other ;  and  -j-b  has  been  removed  from  one 
member  and  appears  as  —b  in  the  other. 

It  is  evident  that  similar  steps  may  be  employed  in  all 
cases.     Hence  we  have  the  following 

Rule. 

Any   term  may  be  transposed  from   one   member  of  an 
equation  to  the  other  by  changing  its  sign. 

It  follows  from  this  that  the  sign  of  every  term  of  an  equa- 
tion may  be  changed;  for  this  is  equivalent  to  transposing 
every  term,  and  then  making  the  first  and  second  members 
change  places.     Thus,  for  example,  suppose  that 
4x  —  8  =  2x  —  16. 


76  SOLUTION    OF  SIMPLE    EQUATIONS. 

Transposing  every  term,  we  have 

—  2x  +  16  =  —  Ax  +  8, 
or  —4a,-  +     8  =  —2x  +  16, 

which  is  the  original  equation  with  the  sign  of  every  term 
changed. 

This  result  can  also  be  obtained  by  multiplying  each  term 
of  the  original  equation  by  —1  (Axiom  3). 

57.  Solution  of  Simple  Equations  with  One  Un- 
known Quantity.  —  Find  the  value  of  x  in  the  equation 
x       18  _  x      x 

2        5    ~  4       5' 

The  least  common  multiple  of  the  denominators  is  20. 
Multiplying  each  term  by  20,  we  get 

10a  —  72  =  hx  —  Ax. 

Transposing  the  unknown  terms  to  the  first  member,  and 
the  known  terms  to  the  second,  we  have 

10a  —  hx  +  Ax  =  72. 
Collecting  the  terms,  we  have 

9a;  =  72. 
Dividing  each  member  by  9  (Axiom  4),  we  have 
x  =  8. 

We  can  now  give  a  general  rule  for  solving  any  simple 
equation  with  one  unknown  quantity. 

Rule. 

Clear  the  equation  of  fractions,  if  necessary;  transpose  all 
the  terms  containing  the  unknown  quantity  to  the  first  member 
of  the  equation,  and  the  knoivn  quantities  to  the  second,  and 
collect  the  terms  of  each  member.  Divide  both  members  by 
the  coefficient  of  the  xinknown  quantity,  and  the  second  member 
is  the  value  required. 


EXAMPLES.  77 


EXAMPLES. 


1 .  Solve  9x  +  35  =  75  +  ox. 

Here  there  are  no  fractions  ;  transposing,  we  have 
9x  —  5x  =  75  —  35. 
Collecting  terms,  Ax  =  40. 

Dividing  by  4,  x  =  10. 

It  is  very  important  for  the  student  to  acquire  the  habit 
of  occasionally  verifying,  that  is,  proving  the  truth  of  his 
results.  The  habit  of  applying  such  proofs  tends  to  con- 
vince the  student,  and  to  make  him  self-reliant  and  confident 
in  his  own  accuracy. 

To  verify  the  result,  in  the  case  of  simple  equations,  we 
substitute  the  value  of  the  unknown  quantity  in  the  original 
equation  ;  if  the  two  members  are  equal  the  result  is  said  to 
be  verified,  or  the  equation  satisfied. 

Thus,  in  the  last  example,  10  is  the  root  of  the  proposed 
equation  (Art.  53).  We  may  verify  this,  i.e.,  we  may  show 
that  x  =  10  satisfies  the  original  equation  by  putting  10  for 
x  in  that  equation. 

Thus  9  x  10  +  35  =     75  +    5  x  10, 

or  90  +  35  =     75  +  50, 

or  125  =  125, 

which  is  clearly  true.     Hence,  since  the  two  members  are 
equal,  x  =  10  satisfies  the  equation. 

2.  Solve  5(x  -  3)  -  7(6  -  a;)  +  3  =  24  -  3(8  -  a;). 
Removing  parentheses, 

5x  _  15  _  42  +  7x  +  3  =  24  -  24  +  3<c. 
Transposing,     5x  +  7x  -  3x  =  24  -  24  +  15  +  42  -  3. 
Collecting  terms,  9x  =  54. 

Dividing  by  9,  x  =  6. 

"We  may  verify  this  result  by  putting  6  for  x  in  the  given 
equation. 

Thus  5(6  -  3)  -  7(6  -  6)  +  3  =  24  -  3(8  -  6), 
or  15  +  3  =  24  -  6, 

or  18  =  18. 


78  EXAMPLES. 

3.    Solve  5x  -  (4.7;  -  7)  (3a  -  5)  =  6  -  3(4a:  -  9)  (a;  -  1). 
Performing  the  multiplications  indicated,  we  have 
hx  -  (12a;2  -  41a:  +  35)  =  6  -  3(4.e2  -  13a;  +  9). 
Removing  the  parentheses,  we  have 

5x  -  12a;2  +  41a;  -  35  =  6  -  12a;2  +  39a;  -  27. 
Erasing  the  term  —12a;2  on  each  side,  and  transposing,  we 
have  5a;  +  41a;  -  39a;  =     6-27  +  35. 

Collecting  terms  7a;  =  14. 

.-.     x  =     2. 
"We  may  verify  this  result  by  putting  2  for  x  in  the  given 
equation. 

The  first  member  becomes 

10  -  (8  -  7)  (6  -  5)  =  10  -  1  =  9  ; 
and  the  second  member  becomes 

6  -  3(8  -  9)  (2  -  1)  =  6  -  3(-l)  =  9. 
Thus,  since  these  two  results  are  the  same,  x  =  2  satisfies 
the  equation. 

Note.  —  In  the  first  line  of  the  solution  of  Ex.  3,  we  did  not 
remove  the  parentheses  until  we  performed  the  multiplications.  The 
beginner  is  recommended  to  put  down  all  his  work  as  full  as  we  have 
done  in  this  example,  in  order  to  insure  accuracy. 

4.  Solve  8a;  -  5 [a?  -  \6  -  5(aj  -  3)f]  =  4a;  +  1. 
Removing  parentheses,  we  have 

8a;  -  5[>  -  21  +  5a;]  =  4aj  +  1, 
8a;  -  5a;  +  105  -  25a;  =  4a;  +  1. 
.-.     -26a;  =  -104. 
x  =  4. 

Solve  the  following  equations  : 

5.  2a;  +  7  =  Bx  +  3.  Ans.  4. 

6.  24a;  -  49  =  19a;  -  14.  7. 

7.  16a;  -  11  =  7x  +  70.  9. 

8.  8(x  -  1)  +  17(a>  -  3)  =  4 (4a;  -  9)  +  4.  3. 

9.  5.x  -  6(.r  -  5)  =  2(.t  +  5)  +  5(a>  -  4).  5. 

10.  8 (a;  -  3)  -  (6  -  2x)  =  2(x  +  2)  -  5(5  -  .r).       3. 

11.  3(169  -  aj)  -  (78  +  aj)  =  29aj.  13. 

12.  7a;  -  39  -  10a;  +  15  =  100  -  33a;  +  26.  5. 


FRACTIONAL   EQUATIONS — EXAMPLES.  79 

13.  118  -  65a;  -123  =  15a;  +  35  -  120a;.  Ans.  1. 

14.  157  -  21  (x  +  3)  =  163  -  15(2a;  -  5).  16. 

15.  97  -  5(.t  +  20)  =  111  -  S(x  +  3).  30. 

16.  x  —  [3  +  \x  —  (3  +  x)\~\  =  5.  5. 

17.  14a;  -  (5a;  -  9)  -  {4  -  3a;  -  (2x  —  3)\  =  30.       2. 

18.  5a;  -  (3a;  -  7)  -  £4  -  2x  -  (6a;  —  3)  |  =  10.       1. 

58.   Fractional  Equations.  —  The  following  are  some 
of  the  most  useful  methods  of  solving  fractional  equations. 

EXAMPLES. 

,      0  ,      5x  -f-  4        7x  +  5       r3       x  —  1 

1.    Solve ! ! —   =5 . 

2  10  5  2 

q3  —  m_  •  the  least  common  multiple  of  the  denominators 

is  10  ;  multiplying  by  10,  we  have 

5  (5a;  +  4)  -  (7a;  -f  5)  =  56-  —  5(a;  —  1) ; 

removing  parentheses, 

25a;  +  20  -  7a;  -  5  =  56  -  5x  +  5  ; 

transposing,  25a;  —  7x  -f-  5i»;  =  56  -+-  5  —  20  +  5 ; 

collecting  terms,  23.«  =  46  ; 

.-.     x  =     2. 

Note.  —  Mistakes  with  regard  to  the  signs  are  often  made  in 
clearing  an  equation  of  fractions.      In  this  equation  the  fraction 

— x  ~     is  regarded  as  a  single  term  with  the  minus  sign  before  It;  it 

is  equivalent  to  —  \{x  —  1).  When  multiplied  by  10,  it  is  well  to  put 
the  result  first  in  the  form  — 5(x  —  1),  and  afterwards  in  the  form 
— 5x  +  5,  in  order  to  secure  attention  to  the  signs. 

2.  Solve  4  -  °L^lA  =  SL  -  I.  Ans.  33. 

8  22        2 

3.  «     i(5aj  +  3)  -  ^(16  -  5a;)  =  37  -  4a:.  6. 

.  Gx  +  15       8a;  —  10       ±x  —  7 


11 


3. 


Note.  —  In  certain  cases  it  is  more  advantageous  not  to  clear  the 
equation  of  fractions  at  once  by  multiplying  it  throughout  by  the  least 
common  multiple  of  the  denominators  (Art.  55),  but  to  clear  it  of 
fractions  partially,  and  then  to  effect  some  reductions,  before  we 
remove  the  remaining  fractions. 


SO  EXAMPLES. 


1        2x  -  3  _  5a;  —  32       a;  +  9 


3  35  9  28 

First  multiply  by  9,  ami  we  have 

3a;  -  12  4-  I8*"27  =  5x  -  32  -  »*  +  81  ; 

T         35  28 

18a;  -  27   .   9a;  +  81       ~         on 
transposing, — 1 — —  =  2.x  —  20. 

Now  clear  of  fractions  by  multiplying  by  the  least  common 
multiple  of  the  denominators,  which  is  5  x  7  X  4,  or  140, 
and  we  get 

72a-  -  108  +  45a;  +  405  =  280a  -  2800 ; 
transposing,  72.x-  +  45a;  -  280a:  =  -2800  +  108  -  405  ; 
collecting  terms,  —  163a;  =  —3097; 

dividing  by  -163,  x  =  19. 

Solve  the  following  equations  : 

6.  ^lA  +  *L+J°  =  5.  Am.  8. 

2  9 

7.  ■+!?  =  3  +  *  16. 

5 

17. 


8. 

x  —  1 

8 

i+2 

+  1 
18 

9. 

4(a;  +  2) 

5 

=  7 

6x 

13 

.0. 

sb  +  20 

9 

3a;  _ 

7 

=  6. 

11     <L=J  +  *J=J+A  =  Q.  4. 

7  3  21 

a-  -f  5        a-  +1         ;b  +  3  1 


12. 


6  9  4 


i3.   ^LzJ)  -  £<«  "  4)  =  \(fl  -  6)  +  A.  6. 

16  12  5  48 


3a;        6  ,      ,    tn^       ,  „,        a;  —  7        ,3 


14.    j  -  ^(x  +  10)  -  (sb  -  3)  -  ~ 


51  4 


4-.  7. 


EQUATIONS    WHOSE   COEFFICIENTS  ARE  DECIMALS.    81 

59.  To  Solve  Equations  -whose  Coefficients  are 
Decimals,  it  is  advisable  generally  to  express  all  the 
decimals  as  common  fractions,  to  insure  accuracy,  and  then 
proceed  as  before  ;  but  it  is  often  found  more  simple  to 
work  entirely  in  decimals. 

EXAMPLES. 

1.  Solve  ,6*a;  +  .25  -  %x  =  1.8  -  .75a  -  f 
Expressing  the  decimals  as  vulgar  fractions,  we  have 

|x  +  i-^=  If  -fa  -i; 
clearing  of  fractions, 

24a,-  +  9  -  Ax  =  68  —  27b  —  12  ; 
transposing,  24a;  —  4a;  +  27a;  =  68  —  12  —  9  ; 
.-.     47a;  =  47  ; 
x  =     1. 

2.  Solve  .375a;  -  1.875  =  .12a;  +  1.185. 
Transposing,  .375a;  -  .12a;  =     1.185  +  1.875; 
collecting  terms,  (.375  -  A2)x  =    3.06; 

that  is  .255a;  =    3.06  ; 

dividing  by  .255,  x  =  12. 

3.  Solve  .5a;  -  .3a;  =  .25a;  -  1.  Ans.  12. 
^1)       "      .2x  -  .16a:  =  .6  -  .3.  8. 

5.       "      2.25a;  -  .125  =  3a;  +  3.75.  -5£. 

60.  Literal  Equations.  —  A  Literal  Equation  is  one  in 
which  some  or  all  the  numbers  are  represented  by  letters. 
Thus 

ax2  +  bx  =  ex  +  4,     and    ax  +  b  =  ex'2  —  cl, 

are  literal  equations.  The  known  numbers  are  usually 
represented  by  the  first  letters  of  the  alphabet,  as  a,  &,  c, 
etc. 

*  .6  denotes  the  repetend  .6666  etc.  =  f .     Similarly  .8  denotes  .888  etc.  =  §. 


82  LITERAL   EQUATIONS — EXAMPLES. 

EXAMPLES. 

1.    Solve  ax  +  h-  =  bx  +  a2. 
Transposing,  we  have 

ax  —  bx  =  a2  —  b2, 
that  is  (a  —  b)x  —  a2  —  b2 ; 

dividing  by  a  —  b,  the  coefficient  of  x,  we  have 
x  =  (a2  -  b2)  -=-  (a  -  b)  =  a  +  b. 

J.    Solve   *  +  -  =  c. 
a       b 

Multiplying  by  ab,  we  have 

Ix  +    ax  =  abc, 

that  is,  (a  +  b)x  =  abc; 

dividing  by  a  -f-  b,  the  coefficient  of  x,  we  have 


a  +  6 
3.   Solve  (a  -  x)  (a  +  x)  =  2a2  +  2ax  -  z2. 

-4ns.  a;  = 


4.  Solve  2a;  4-  6as  -  a  =  Sx  -  2c.  x  =  a  ~  2c. 

6  -  1 

5.  Solve  ax  —  &a;  -f  b2  =  a2.  .*•  =  a  -+-  &. 

6.  Solve  (a  +  »)(&  +  a-)  =  a(&  +  c)  +*—  +  a2. 

Ans.  x  =  — . 
6 

7.  Solve  a(x  —  a)  +  b(x  -  &)  =  2a&.  .<■  =  «  +  6. 
.s.  Solve  2(sb  -  a)  +  3(as  -  2a)  =  2a.  x  =  2a. 
:».    Solve  }(.<•  +  a  +  &)  +  K*  4-  a  -  6)  =  A. 

^rl»s.  x  =  b  —  a. 
10.   Solve  (a  +  ba?)(6  +  aa?)  =  a&(a£  -  1). 


x  = 


PROBLEMS   LEADING    TO   SIMPLE   EQUATIONS.  83 

61.  Problems    Leading    to    Simple    Equations. — 

The  preceding  principles  may  now  be  employed  to  solve 
various  problems. 

A  Problem  is  a  question  proposed  for  solution.  In  a 
problem  certain  quantities  are  given  or  known,  and  certain 
others  which  have  some  assigned  relations  to  these,  are 
required. 

A  Theorem  is  a  truth  requiring  proof. 

Axioms,  Problems,  and  Theorems,  are  called  Propositions. 

The  Solution  of  a  problem  by  Algebra  consists  of  two  dis- 
tinct parts  :  (1)  The  Statement  of  the  problem,  and  (2)  the 
Solution  of  the  equation  of  the  problem. 

The  Statement  of  the  problem  is  the  process  of  expressing 
the  conditions  of  the  problem  in  Algebraic  language  by  an 
equation.  The  statement  of  the  problem  is  often  more 
difficult  to  beginners  than  the  solution  of  the.  equation.  No 
rule  can  be  given  for  the  statement  of  every  particular 
problem.  Much  must  depend  on  the  skill  of  the  student, 
and  practice  will  give  him  readiness  in  this  process.  The 
following  is  the  general  plan  of  finding  the  equation  : 

1.  Study  the  problem,  to  ascertain  what  quantities  in  it  are 
known  and  what  are  unknown,  and  to  understand  it  fully,  so 
as  to  be  able  to  prove  the  correctness  or  incorrectness  of  any 
proposed  answer. 

2.  Represent  the  unknown  quantity  by  one  of  the  fined 
letters  of  the  alphabet,  say  x,  and  express  in  Algebraic 
language  the  relations  which  hold  betiveen  the  known  and  un- 
known quantities;  an  equation  will  thus  be  obtained  tohich  can 
be  solved  by  the  methods  already  given,  and  from  which  the 
value  of  the  unknoivn  quantity  may  be  found. 

Note  1.  —  Problems  may  often  involve  several  unknown  quantities, 
but  in  the  present  chapter  we  shall  consider  only  problems  in  which 
there  is  one  unknown  quantity,  or  in  which,  if  there  are  several,  they 
are  so  related  to  one  another  that  they  can  all  be  expressed  in  terms 
of  some  one  of  them. 


84  EXAMPLES. 

EXAMPLES. 

1 .    What  number  is  that  whose  double  exceeds  its  half  by 
27? 

Let  x  represent  the  number  ; 
then  2x  represents  the  double  of  the  number, 

and  ^  represents  the  half  of  the  number. 

Since   from   the    conditions   of    the   problem    the   double 
exceeds  the  half  by  27,  we  have  for  the  equation 


Clearing  of  fractions, 

Ax  —  x  —  54, 

that  is,  3aj  =  54. 

.-.  x  =18. 

Hence  the  required  number  is  18. 

18 
Verification,  2x18 r-  =  27. 

2.    The  sum  of  two  numbers  is  28,  and  their  difference  is 
4  ;  find  the  numbers. 

Let  x  =  the  smaller  number  ; 
then  x  +  4  =  the  greater  number ; 

and  since,  from  the  conditions  of  the  problem,  the  sum  is  to 
be  equal  to  28,  we  have  for  the  equation 
x  +  x  +  4  =  28 ; 
that  is  2x  =  24. 

.-.     x  =  12, 
and  X  +  4  =  1(1, 

so  that  the  numbers  are  12  aud  1G. 

Verification,  16  +  12  =  28,  and  16  —  12  =  4. 

The  beginner  is  advised  to  test  each  solution  by  proving 
that  it  satisfies  the  data  of  the  question. 

8.    A  has  $80  and  B  has  $15.      IIow  much  must  A  give  to 
B  in  order  that  he  may  have  just  four  limes  as  mueh  as  B? 


EXAMPLES.  85 

Let  x  =  the  number  of  dollars  that  A  gives  to  B  ; 

then  80  —  a;  =  the  number  of  dollars  that  A  has  left, 
and  15  +  x  =  the  number  of  dollars  that  B  will  have  after 
receiving  x  dollars  from  A. 

But  A  has  now  four  times  as  much  as  B  ;  hence  we  have 
the  equation 

80  -  x  =  4(15  +  x), 
that  is,  80  —  x  =  60  +  4a-, 

transposing  and  uniting,  —  hx  —  —20, 
dividing  by  —5,  x  =  4. 

Hence  A  must  give  $4  to  B. 

4.  A  father  is  six  times  as  old  as  his  son,  and  in  four 
years  he  will  be  four  times  as  old.     How  old  is  each? 

Let         x  —  the  son's  age  in  years, 
then  Gx  =  the  father's  age  in  years. 

Also  x  +  4  =  the  son's  age  in  years,  after  four  years, 
and  6x  +  4  =  the  father's  age  in  years,  after  four  years. 
Hence,  from  the  conditions,  we  have  the  equation 

6x  +  4  =  4(»  +  4), 
that  is, '  Gx  +  4  =  '4a>  +  16; 

.-.     x  =  6,  the  son's  age, 
and  6.t  =  36,  the  father's  age. 

5.  Divide  60  into  two  parts,  so  that  3  times  the  greater 
may  exceed  100  by  as  much  as  8  times  the  less  falls  short  of 
200. 

Let  x  =  the  greater  part, 

then  60  —  x  =  the  less. 

Also  3a;  —  100,=  the  excess  of  3  times  the  greater  over 
-100,  and  200  —  8(60  —  aj)  =  the  number  that  8  times  the 
less  falls  short  of  200.  Hence,  from  the  conditions,  we 
have  the  equation 

3.c  -  100  =  200  -  8(60  -  x), 
that  is,  3a;  -  100  =  200  -  480  +  8a;, 

hence,  —ox  =  —180. 

.-.    x  =  36,  the  greater  part. 
60  —  x  =  24,  the  less. 


8b  EXAMPLES. 

G.  A  line  is  2  feet  4  inches  long ;  it  is  required  to  divide 
it  into  two  parts,  such  that  one  part  may  be  three-fourths  of 
the  other  part. 

Let  x  =  the  number  of  inches  in  the  larger  part, 
then  fa  =  the  number  of  inches  in  the  other  part. 
Hence,  from  the  conditions,  we  have  the  equation 

x  +  -lx  =  28, 
that  is,  4x  +  3x  =  112. 

.-.     x  =  16. 
Thus  oue  part  is  16  inches  long,  and  the  other  part  12  inches 
long. 

7.  Divide  $47  between  A,  B,  C,  so  that  A  may  have  $10 
more  than  B,  and  B  $8  more  than  C. 

Note  2.  —  Here  there  are  really  three  unknown  quantities,  but  it 
is  only  necessary  to  represent  the  number  of  dollars  the  last  has  by  a 
symbol. 

Let  x  =  the  number  of  dollars  that  C  has, 

then  x  4-  8  =  the  number  of  dollars  that  B  has, 

and    x  +  8  +  10  =  the  number  of  dollars  that  A  has. 
Hence  we  have  the  equation 

x  +  (a;  -j-  8)  +  (aJ  +  8  +  10)  =  47, 
.-.      3a;  =  21, 
x  =  7; 
so  that  C  has  $7,  B  $15,  A  $25. 

8.  A  person  spent  £28.  4,s.  in  buying  geese  and  ducks  ;  if 
each  goose  cost  7s.,  and  each  duck  cost  3s.,  and  if  the  total 
number  of  birds  bought  was  108,  how  many  of  each  did  he 
buy? 

Note  3. —In  questions  of  this  kind  it  is  of  essential  importance 
to  have  all  concrete  quantities  of  the  same  kind  expressed  in  the  same 
denomination;  in  the  present  instance  it  will  be  convenient  to  express 
the  money  in  shillings.  In  Ex.  <i  it  was  convenient  to  express  the 
length  in  inches. 

Let  x  =  the  number  of  geese, 

then        108  —  x  =  the  number  of  ducks. 
Also  7.c  =  the  number  of  shillings  the  geese  cost, 

and  :J(10H  —  x)  =  the  Dumber  of  shillings  the  ducks  eost. 


EXAMPLES.  87 

But  from  the  conditions  of  the  question,  the  whole  cost  of 
the  geese  and  ducks  is  £28.  4s.,  i.e.,  564  shillings.  Hence 
we  have  the  equation 

7x  +  3(108  —  x)  =  564, 
that  is,    Ix  -f-  324  —  3x  =  564, 

.•.     x  =  60,  the  number  of  geese, 
and  108  —  x  =  48,  the  number  of  ducks. 

9.  A  can  do  a  piece  of  work  in  12  hours,  which  B  can  do 
in  4  hours.  A  begins  the  work,  but  after  a  time  B  takes  his 
place,  and  the  whole  work  is  finished  in  6  hours  from  the 
beginning.     How  longVdid  A  work  ? 

Let  x  =  the  number  of  hours  that  A  worked, 

then      6  —  x  =  the  number  of  hours  that  B  worked. 
Also  -^  =  the  part  A  does  in  1  hour,  since  he  can  do 

the  whole  work  in  12  hours. 

Therefore  —  =  the  part  done  by  A  altogether. 

Also  \  =  the  part  B  does  in  1  hour,  since  he  can  do 

the  whole  work  in  4  hours. 
Therefore  |(6  —  x)  =  the  part  done  by  B  altogether. 

But  A  and  B  together  do  the  whole  work  ;  hence  the  sum 
of  the  parts  of  the  work  that  they  do  separately  must  equal 
unity;  and  we  have  for  the  equation 

^-+1(6  -x)  =  1. 
12        4V  ' 

Multiplying  by  12,  we  have 

x  +  3(6  -  x)  =     12, 

.-.      -2x  =  -6. 

x  =       3. 

Hence  A  worked  for  3  hours. 

Note  4.  —  It  should  be  remembered  that  x  must  always  represent 
a  number;  what  is  called  the  unknown  quantity  is  really  an  unknown 
number.  In  the  above  examples  the  unknown  quantity  x  represents 
a  number  of  dollars,  years,  inches,  etc.  For  instance,  in  Ex.  6,  we  let 
x  denote  the  number  of  inches  in  the  longer  part;  beginners  often  say, 


88  EXAMPLES. 

"  let  x  =  the  longer  part,"  or,  "  let  x  =  a  part,"  which  Is  not  definite, 
because  a  part  may  be  expressed  in  various  ways,  in  feet,  or  inches,  or 
yards.  Again,  in  Ex.  7,  we  let  x  =  the  number  of  dollars  that  C  has-, 
beginners  often  say,  "let  x  =  C's  money,"  which  is  not  definite, 
because  C's  money  may  be  expressed  in  various  ways,  in  dollars,  or  in 
pounds,  or  as  a  fraction  of  the  whole  sum.  The  student  must  be 
careful  to  avoid  beginning  a  solution  with  a  vague  and  inexact 
statement. 

It  may  seem  to  the  student  that  some  of  the  problems  which  are 
given  for  exercise  can  be  readily  solved  by  Arithmetic,  and  he  may 
therefore  be  inclined  to  undervalue  the  power  of  Algebra  and  consider 
it  unnecessary.  We  may  remark,  however,  that  by  Algebra  the  student 
is  enabled  to  solve  all  the  problems  given  here,  without  any  uncer- 
tainty; and  also,  he  will  find  as  he  proceeds,  that  he  can  solve 
problems  by  Algebra,  which  would  be  extremely  difficult  or  entirely 
impracticable,  by  Arithmetic  alone. 

10.  The  difference  between  two  numbers  is  8  ;  if  2  be 
added  to  the  greater  the  result  will  be  three  times  the 
smaller:  find  the  numbers.  Ans.  13,  5. 

11.  A  man  walks  10  miles,  then  travels  a  certain  distance 
by  train,  and  then  twice  as  far  by  coach.  If  the  whole 
journey  is  70  miles,  how  far  does  he  travel  by  train  ? 

Ans.  20  miles. 

12.  What  two  numbers  are  those  whose  sum  is  58,  and 
difference  28?  .4ns.  15,  43. 

13.  If  288  be  added  to  a  certain  number,  the  result  will 
be  equal  to  three  times  the  excess  of  the  number  over  12  : 
find  the  number.  Ans.  162. 

14.  Find  three  consecutive  numbers  whose  sum  shall 
equal  84.  Ans.  27,  28,  29.. 

15.  Find  two  numbers  differing  by  10.  whose  sum  is  equal 
to  twice  their  difference.  Ans.  15,  5. 

Hi.  Find  a  number  such  that  if  5,  15,  and  35  are  added 
to  it,  the  product  of  the  first  and  third  results  may  be  equal 
to  the  square  of  the  second.  Ans.  5. 

*  17.  A  is  twice  as  old  as  B,  and  seven  years  ago  their 
united  ages  amounted  to  as  many  years  as  now  represent  the 
age  of  A  :    find  the  ages  of   A  and  15.  Ans.  2«,  11. 


EXAMPLES. 


EXAMPLES. 


89 


Solve  the  following  equations  : 

1.  3oj  +  15  =  x  +  25.  -4tt*.  5. 

2.  2a;  +  3  =  16  -  (2x  -  3).  4. 

3.  7(25  -  x)  -  2x  =  2(3x  -  25).  15. 

4.  5a;  -  17  +  3a;  -  5  =  6a;  -  7  -  8a;  +  H5.  13. 

5.  5(i»  +  2)  =  3(a!  +  3)  +  1.  0. 

6.  2(x  -  3)  =  5(a)  >  1)  +  2x  -  1.  -2. 

7.  2(x  -  1)  -  3  (a;  -  2)  +  4 (a  -  3)  +  2  =  0.  2. 

8.  5a;  +  6(3!  +  1)  -  7(x  +  2)  -  8 (a  +  3)  =  0.       -8. 

9.  (a  +  i)(2as  +  1)  =  (a?  +  8)(2as  +  3)  -  14.  1. 

10.  (x+l)2-(a;2-l)=a;(2a;+l)  -2(«+2)  (x-4-1)  +20. 

Ans.  2. 

11.  6(«2-3x+2)-2(.i-2-l)=4(a;+l)(a;+2;)-24.         1. 

12.  2a;  -  5^3a;  -  7  (4a;  -  9)\  =  66.  3. 

13.  3(5  -  6x)  -  5[>  -  5|1  -  3(aj  -  5)|]  =  23.       4. 

14.  (a;  +  l)2  +  2(3!  +  3)2  =  3x(x  +  2)  +  35.  2. 

15.  84+(s+4)(s-8)(a+5)  =  (aj+l).(*+2)(»+S).     1. 

16.  (a  +  1)  (a;  +  2)  (a;  +  6)  =  x3  +  9x2  +  4(73!  -  1) .       2. 

17  *  _  ?.  =  1.                                                               -20. 
5       4 

18.  5^zi  +  *Zli  =  3. 

19.  1(3!   4-   1)   -  f(3!   -   1)   = 

20.  A(2  -  3!)  -  i(5.r  +  21)  = 

21-  ~2~~  +       3       +       4 

22.  *^J?  -  x~  4  =  g~3  -  (3!  -  2).                      2*. 
2                3                2            V 


5. 

-5. 

a;  +  3. 

912 

8  =  0. 

-9A 

±_i  _  2-c-  i  +  ii  =  0.  -16. 

2  5 

3x  +  5        21 


23-    "1 5~ 


24. 


25.    Z*  -  T\(x  -  11)  =  f(x-  -  25)  +  34.  25. 

5 


90  EXAMPLES. 


26.    i(.,  _  8)  +  :L±^  +  --=-^-  =  7  -    23  -  * 

4  7 


27. 


—  (•—^io1)- *<■—"> 


a. 


28.  1^-1^  +  13  =  0.  x. 

3  G 

29.  !L±I  _  ^J^i  +  4X  =  12  +  2a?  -  1.  3. 

3  4  6 

30.  *±1  _  ^|J  =  b^l_ZL£J  +  i.  28. 

31.  §«LZLl  _  13  ~  *  =  25  _  ¥(.,  +  3).  2. 

o  2  o 

32.  IZL*  +  *-lJ>  +  1^  +  LlL?  +  f  =  0.  4. 

o  4  o  b 

33.  5a?~  3  -  "LzLi;  =  ^  _|_  i9(it;  _  4).  2. 

7  3  2         ^  V  ; 

34.  p  4-  .25a;  -  .Sx  =  x  -  3.  12. 

35.  J)x  —  .2a;  =  .3a;  —  1.5.  27. 

3G.    1.5  =  ^  -  -09a?  ~  -18. 
.2  .9 

37.  (a  +  &)»  +  (a  -  b)x  =  <r. 

38.  (a  +  &)rc  4-  (6  -  a)aj  ==  &3. 

39.  ■§(«  +  ,-)  4-  l(2a  4-  ,•)  4-  \(3a  4-  «)  =  3a. 


a 
2 

h 
2' 
a. 

An     xa    .    xb  -,,.. 

40.  —  4 =  a-  4-  lr.  „b. 

b         a 

41.  (a2  4-  a?)(62  4-  x)  =  {ub  4-  a?)3.  0. 

42.  a(as  4-  a)  +  &(&  -  x)  =  2a&.  &  -  a. 

43.  aa!(aj  4-  a)  +  &a;(a:  4-  b)  =  (a  4-  6)  (.>•  4-  a)  (x  4-  &). 

J//.S-.  —  $(a  4-  6). 

44.  (a?-a)8+(a!-6)»+(aj-c)8=3(aj-a)(aj-6)(a;-c). 

Ans.  l(a  4-6  4-  c). 

15.    Twenty-three  times  a  certain  Dumber  is  as  much  above 

11  as  16  is  above  seven  times  the  number:  liml  it.      J/**.  1. 


EXAMPLES.  91 

46.  Divide  105  into  two  parts,  one  of  which  diminished 
Irv  20  shall  be  equal  to  the  other  diminished  by  15. 

Ans.  50,  55. 

47.  The  sum  of  two  numbers  is  8,  and  one  of  them  with 
22  added  to  it  is  five  times  the  other :  find  the  numbers. 

l  Ans.  3,  5. 

48.  A  and  B  begin  to  play  each  with  $60.  If  they  play 
till  A's  money  is  double  B's,  what  does  A  win?       Ans.  $20. 

49.  The  difference  between  the  squares  of  two  consecutive 
numbers  is  121  :  find  the  numbers.  Ans.  60,  61. 

50.  Divide  $380  between  A,  B,  and  C,  so  that  B  may 
have  $30  more  than  A,  and  C  may  have  $20  more  than  B. 

Ans.  A  $100,  B  $130,  C  $150. 

51.  A  father  is  four  times  as  old  as  his  son  ;    in  24  years  f 
he  will  only  be  twice  as  old  :  find  their  ages.       Ans.  48,  12. 

52.  A  is  25  years  older  than  B,  and  A's  age  is  as  much 
above  20  as  B's  is  below  85  :  find  their  ages.      Ans.  65,  40. 

53.  The  sum  of  the  ages  of  A  and  B  is  30  years,  and  five  y 
years  hence  A  will  be  three  times  as  old  as  B :    find  their 
present  ages.  Ans.  25,  5. 

54.  The  length  of  a  room  exceeds  its  breadth  by  3  feet ; 
if  the  length  had  been  increased  by  3  feet,  and  the  breadth 
diminished  by  2  feet,  the  area  would  not  have  been  altered : 
find  the  dimensions.  Ans.  15  ft.,  12  ft. 

55.  There  is  a  certain  fish,  the  head  of  which  is  9  inches 
long ;  the  tail  is  as  long  as  the  head  and  half  the  body  ;  and 
the  body  is  as  long  as  the  head  and  tail  together  :  what  is  the 
length  of  the  fish?  Ans.  6  ft. 

56.  The  sum  of  $76  was  raised  by  A,  B,  and  C  together ; 
B  contributed  as  much  as  A  and  $10  more,  and  C  as  much 
as  A  and  B  together :  how  much  did  each  contribute  ? 

Ans.  $14,  $24,  $38. 

57.  After  34  gallons  had  been  drawn  out  of  one  of  two 
equal  casks,  and  80  gallons  out  of  the  other,  there  remained 
just  three  times  as  much  in  one  cask  as  in  the  other :  what 
did  each  cask  coutain  when  full?  Ans.  103. 


92  EXAMPLES. 

58.  Divide  the  number  20  into  two  parts  such  that  the 
sura  of  three  times  one  port,  and  live  times  the  other  part, 
may  be  84.  Ans.  8,  12. 

59.  A  person  meeting  a  company  of  beggars  gave  4  cents 
to  each,  and  had  16  cents  left;  he  found  that  he  should 
have  required  12  cents  more  to  enable  him  to  give  the 
beggars  6  cents  each  :    how  many  beggars  were  there  ? 

Ans.  14. 

60.  Divide  100  into  two  parts  such  that  if  a  third  of  one 
part  be  subtracted  from  a  fourth  of  the  other,  the  remainder 
may  be  11.  Ans.  24,  76. 

61.  Divide  60  into  two  parts  such  that  the  difference 
between  the  greater  and  64  may  be  equal  to  twice  the 
difference  between  the  less  and  38.  Ans.  36,  24. 

62.  Find  a  number  such  that  the  sum  of  its  fifth  and  its 
seventh  shall  exceed  the  sum  of  its  eighth  and  its  twelfth  by 
113.  Ans.  840. 

63.  An  army  in  defeat  loses  one-sixth  of  its  number  in 
killed  and  wounded,  and  4000  prisoners  ;  it  is  re-enforced 
by  3000  men,  but  retreats,  losing  one-fourth  of  its  number 
in  doing  so  ;  there  remain  18000  men  :  what  was  the  original 
force?  Am.  30000. 

64.  One-half  of  a  certain  number  of  persons  received  18 
cents  each,  one-third  received  24  cents  each,  and  the  rest 
received  30  cents  each ;  the  whole  sum  distributed  was 
$5.28  :  how  many  persons  were  there?  Ans.  24. 

65.  A  father  has  six  sons,  each  of  whom  is  four  years 
older  than  his  next  younger  brother ;  and  the  eldest  is  three 
times  as  old  as  the  youngest :  find  their  respective  ages. 

Ans.  10,  14,  18,  22,  26,  30. 

66.  A  man  left  his  property  to  be  divided  between  his 
three  children  in  such  a  way  that  the  share  of  the  eldest  was 
to  be  twice  that  of  the  second,  and  the  share  of  the  second 
twice  that  of  the  youngest ;  it  was  found  that  the  eldest 
received  $3000  more  than  the  youngest:  how  much  did  each 
receive?  Ans.  §4000,  §2000,  §1000. 


'  EXAMPLES.  93 

67.  A  sum  of  money  is  divided  among  three  persons  ;  the 
first  receives  $10  more  than  a  third  of  the  whole  sum ; 
the  second  receives  $15  more  than  a  half  of  what  remains  ; 
and  the  third  receives  wha\  is  over,  which  is  $70  :  find  the 
original  sum.  Ans.  $270. 

68.  In  a  cellar  one-fifth  of  the  wine  is  port  and  one-third 
claret ;  besides  this  it  contains  15  dozen  of  sherry  and  30 
bottles  of  spirits  :  how  much  port  and  claret  does  it  contain  ? 

Ans.  90  port,  150  claret. 

69.  Two-fifths  of  A's  money  is  equal  to  B's,  and  seven- 
ninths  of  B's  is  equal  to  C's  :  in  all  they  have  $770,  what 
have  they  each?  Ans.  A  $450,  B  $180,  C  $140. 

70.  A,  B,  and  C  have  $1285  between  them  ;  A's  share  is 
greater  than  five-sixths  of  B's  by  $25,  and  C's  is  four- 
fifteenths  of  B's  :   find  the  share  of  each. 

Ans.  A  $525,  B  $600,  C  $160. 

71.  The  width  of  a  room  is  two-thirds  of  its  length;  if 
the  width  had  been  3  feet  more,  and  the  length  3  feet  less, 
the  room  would  have  been  square  ;  find  its  dimensions. 

Ans.  12  ft.,  18  ft. 

72.  A  man  left  $2200  to  be  divided  among  his  four 
children  A,  B,  C,  D  ;  of  whom  B  was  to  have  twice  as  much 
as  A,  C  as  much  as  A  and  B  together,  and  D  as  much  as  C 
and  B  together  :  how  much  had  each  ? 

Ans.  $200,  $400,  $600,  $1000. 

73.  A  sum  of  money  is  to  be  distributed  among  three 
persons,  A,  B,  and  C ;  the  shares  of  A  and  B  together 
amount  to  $240  ;  those  of  A  and  C  to  $320  ;  and  those  of  B 
and  C  to  $368 :  find  the  share  of  each  person. 

Ans.  $96,  $144,  $224. 

74.  Two  persons  A  and  B  are  travelling  together  ;  A  has 
$100,  and  B  has  $48  ;  they  are  met  by  robbers  who  take 
twice  as  much  from  A  as  from  B,  and  leave  to  A  three  times 
as  much  as  to  B  :  how  much  was  taken  from  each? 

Ans.  $88,  $44. 

75.  A  wine  merchant  has  two  sorts  of  wines,  one  sort 


04  EXAMPLES. 

worth  $2.40  a  quart,  and  the  other  worth  $1  a  quart ;  from 
these  he  wants  to  make  a  mixture  of  100  quarts  worth  $2.80 
a  quart :  how  many  quarts  must  he  take  from  each  sort  ? 

Ans.  75,  25. 

76.  In  a  mixture  of  wine  and  water  the  wine  composed 
25  gallons  more  than  half  of  the  mixture,  and  the  water  5 
gallons  less  than  a  third  of  the  mixture  :  how  man}'  gallons 
were  there  of  each?  Ans.  85,  35. 

77.  In  a  lottery  consisting  of  10000  tickets,  half  the 
number  of  prizes  added  to  one-third  the  number  of  blanks 
was  3500  :  how  many  prizes  were  there  in  the  lottery? 

Ans.  1000. 

78.  In  a  certain  weight  of  gunpowder  the  saltpetre  com- 
posed 6  lbs.  more  than  a  half  of  the  weight,  the  sulphur 

5  lbs.  less  than  a  third,  and  the  charcoal  3  lbs.  less  than  a 
fourth :  how  many  lbs.  were  there  of  each  of  the  three 
ingredients?  Ans.  18,  3,  3. 

79.  A  general,  after  having  lost  a  battle,  found  that  he 
had  left  fit  for  action  3600  men  more  than  half  of  his  army  ; 
600  men  more  than  one-eighth  of  his  army  were  wounded ; 
and  the  remainder,  forming  one-fifth  of  the  arm}',  were  slain, 
taken  prisoners,  or  missing  :  what  was  the  number  of  the 
army?  Am.  24000. 

80.  How  many  sheep  must  a  person  buy  at  $28  each  bo 
that  after  paying  20  cents  a  score  for  folding  them  at  night 
he  may  gain  $319.20  by  selling  them  at  $32  each  ?     Ans.  80. 

81.  A  certain  sum  of  money  was  shared  among  five 
persons,  A,  B,  C,  D,  and  E ;  B  received  $10  less  than  A ; 
C  received  $16  more  than  B;  D  received  $5  less  than  C; 
and  E  received  $15  more  than  D  ;  and  it  was  found  that  E 
received  as  much  as  A  and  B  together :  how  much  did  each 
receive?  Ans.  $26,  $16,  $32,  $27.  $12. 

82.  A  tradesman  starts  with  a  certain  sum  of  money  ;  at 
the  end  of  the  first  year  he  had  doubled  his  original  stock, 
all  but  $100;  also  at  the  end  of  the  second  year  he  had 
doublet!  the  stock  at  the  beginning  of  the  second  year,  all 


EXAMPLES.  9") 

but  $100;  also  in  like  manner  at  the  end  of  the  third  year; 
and  at  the  end  of  the  thfyd  year  he  was  three  times  as  rich 
as  at  first :  find  his  original  stock.  Ans.  $140. 

83.  A  person  went  to  a  tavern  witli  a  certain  sum  of 
money  ;  there  he  borrowed  as  much  as  he  had  about  him,  and 
spent  a  dollar  out  of  the  whole  ;  with  the  remainder  he  went 
to  a  second  tavern,  where  he  borrowed  as  much  as  he  had 
left,  and  also  spent  a  dollar ;  and  he  then  went  to  a  third 
tavern,  borrowing  and  spending  as  before,  after  which  he 
had  nothing  left :  how  much  had  he  at  first?    Ans.  87£  cents. 

84.  A  crew  which  can  pull  at  the  rate  of  nine  miles  an 
hour,  finds  that  it  takes  twice  as  long  to  come  up  a  river  as 
to  go  down  :  at  what  number  of  miles  an  hour  does  the  river 
flow?  Ans.  3. 

85.  A  and  B  play  at  a  game,  agreeing  that  the  loser  shall 
always  pay  to  the  winner  $1  more  than  half  the  money  the 
loser  has  ;  they  commence  with  equal  sums  of  money,  but 
after  B  has  lost  the  first  game  and  won  the  second,  he  has 
twice  as  much  as  A  :  how  much  had  each  at  the  commence- 
ment? Ans.  $6. 

86.  A  and  B  find  a  purse  with  dollars  in  it.  A  takes  out 
$2  and  one-sixth  of  what  remains  ;  then  B  takes  out  $3  and 
one-sixth  of  what  remains  :  and  then  they  find  that  they 
have  taken  out  equal  shares.  How  many  dollars  were  in  the 
purse,  and  how  many  did  each  take? 

Ans.  $20  in  the  purse  ;  $5  taken  by  each. 


9G      WHEN  ALL    TERMS  HAVE   ONE   COMMON  FACTOR. 


CHAPTER    VII. 

FACTORING  — GREATEST    COMMON    DIVISOR- 
LEAST     COMMON     MULTIPLE. 

62.  Definitions.  —  Factoring  is  the  process  of  resolving 
a  quantity  into  its  factors. 

The  Factors  of  a  quantity  are  those  quantities  which 
multiplied  together  produce  it.  A  factor  of  a  quantity  is 
therefore  a  divisor  of  the  quantity,  i.e.,  it  wiU  divide  the 
quantity  without  a  remainder.  Thus,  a  is  a  factor  or  divisor 
of  abc,  and  b  is  a  factor  or  divisor  of  ab  —  b2. 

Note.  —In  Division  (Chap.  V.)  we  had  given  the  product  of  two 
factors  and  one  of  the  factors,  and  we  showed  how  to  find  the 
other  factor.  In  the  present  chapter  we  shall  consider  cases  in  which 
the  factors  of  an  expression  can  be  found  when  none  of  the  factors 
are  given. 

A  Prime  Quantity  is  one  which  has  no  integral  factor 
except  itself  and  unity.  Thus,  a,  b,  and  a  +  c  are  prime 
quantities  ;  while  ab,  and  ac  +  be  are  not  prime. 

Quantities  are  said  to  be  prime  to  each  other  or  relatively 
prime,  when  unity  is  the  only  integral  factor  common  to 
both.     Thus,  ah  and  cd  are  prime  to  each  other. 

A  Composite  Quantity  is  one  which  is  the  product  of  two 
or  more  integral  factors,  neither  of  which  is  unity  or  the 
quantity  itself.  Thus,  ax  +  x2  is  a  composite  quantity, 
the  factors  of  which  are  x  and  a  +  x. 

63.  When  All  the  Terms  have  one  Common 
Factor.  —  When  each  term  of  a  polynomial  is  divisible 
by  a  common  factor,  the  polynomial  may  be  simplified  by 
the  following 


EXAMPLES.  97 

Rule. 

Divide  each  term  of  the  polynomial  separately  by  the  com- 
mon factor,  and  enclose  the  quotient  within  parentheses,  the 
common  factor  being  placed  outside  as  a  coefficient;  then  the 
divisor  will  be  one  factor  and  the  quotient  the  other. 

EXAMPLES. 

1 .  Factor  the  expression  3a2  —  6a6. 

Here  we  see  that  the  terms  have  a  common  factor,  da ; 
therefore,  dividing  the  polynomial  by  3a,  we  obtain  for  the 
quotient  a  —  26.  Hence  the  two  factors  are  3a  and  a  —  2b. 
.-.     3a2  -  Gab  =  3a(a  -  26). 

Similarly 

2.  5a2bx*  -  lbab2x3  -  20absxi  =  5abx3(ax  -  36  -  462.r). 

Factor  the  following  expressions  : 

o 


3.    x2  —  ax. 

Ans.  x{x  —  a). 

4.    x3  —  x2. 

x2(x  -  1). 

5.    a2  -  ab2. 

a(a  -  b2). 

(§   8x  -  2a;2. 

2.t-(4  -  x). 

.7.    bax  —  ba3x2. 

bax  (I  —  a2x). 

8.    x3  —  x2y. 

x2(x  —  y). 

9.    bx  -  2bx2y. 

bx  (1  —  bxy). 

10.    16a;2  +  6ix2y. 

16a;2(l  +  Ay). 

11.    54  -  81a. 

27(2  -  3a). 

12.    3,<-3  —  Gar  +  9x. 

3.r(a;2  -  2»  +  3) . 

13.    6a2bx3  +  2ab2xi  +  4a6ar\ 

2a6«8(3a  +  6a;  +  2a;2). 

l4\    72x2y  —  84:xy2  +  60ar>2. 

12a;?/ (6a;  —  ly  +  bxy). 

64.  Expressions  containing  Four  Terms.  —  When  a 

polynomial  contains  four  terms  which  can  be  arranged  in 
pairs  that  have  a  common  binomial  factor,  the  polynomial 
may  be  simplified  by  the  following 

Rule. 

Divide  the  polynomial  by  the  common  binomial  factor ;  then 
the  divisor  will  be  one  factor  and  the  quotient  the  other. 


98  EXAMPLES. 


EXAMPLES. 


1.  Resolve  into  factors  x2  —  ax  +  bx  —  ab. 

Here  we  see  that  the  first  two  terms  contain  a  factor  x,  and 
the  last  two  terms  a  factor  b  ;  therefore  we  factor  the  first 
two  and  last  two  terms  by  Art.  G3,  and  obtain  x(x  —  a) 
and  b (x  —  a).  We  now  see  that  the  two  pairs  have  the 
common  binomial  factor  x  —  a.  Dividing  by  x  —  a  we 
obtain  the  quotient  x  +  b  for  the  other  factor. 
The  work  therefore  will  stand  as  follows  : 

x2  —  ax  +  bx  —  ab  =  x(x  —  a)  +  b(x  —  a) 
=  (x  -  a)(x  +  b). 

2.  Resolve  into  factors  6x2  —  9ax  +  46a;  —  Gab. 

6x2  -  Sax  +  Abx  -  Gab  =  3x(2x  -  3a)  +  2b(2x  -  3a) 
=  (2x  -  3a)  (Bx  +  2b). 

3.  Resolve  into  factors  12a2  —  4«6  —  3aar  +  &02. 
12a2  -  4a&  -  3aar  +  frr  =  4a  (3a  -  6)  -  .r(3a  -  6) 

=  (3a  -  b)  (4a  -  x2). 

Note.  —  It  is  not  necessary  always  to  factor  in  the  same  way.  In 
the  first  line  of  work  it  is  usually  sufficient  to  see  that  each  pair 
contains  some  common  factor;  and  any  suitably  chosen  pairs  will 
bring  out  the  same  result.  Thus,  in  the  last  example,  we  may  have 
a  different  arrangement,  and  enclose  the  first  and  third  terms  in  one 
pair,  and  the  second  and  fourth  in  another  as  follows : 

12a2  -  4ab  -  3ax2  +  bx2  =  \'2<i2  —  Sax2  —  (-lab  -  />.<•-) 
_  3rt(4a  —  x2)  -  6(4a  -  xz) 
-  (4a  _  x2)  (3a  -  b), 
which  is  the  same  result  as  before. 

Resolve  into  factors 

4.  a2  +  ab  +  ac  +  be.  Ans.  (a  +  b)(a  +  c) 

f>.  a2  -  ac  +  ab  -  be.  (a  -  c)(a  +  6) 

<;.  (/-r-  +  oca*  +  abc  +  bd.  (ac  +  d)(ac  +  b) 

7.  a2  +  3a  +  ac  +  3c.  (a  +  3)  (a  +  c) 

8.  2ax  +  a?/  +  2&aj  +  by.  (2x  +  2/)  (a  +  6) 

9.  3ax  -  bx  -  3a?/  +  &//.  (3a  -  &)  (•'*  ~  SO 

10.  a.i;2  +  6x2  +  2a  +  26.  (a  +  b){x2  +  2) 

11.  x1  -  3x  -  a#  +  8y.  (a  -  3)  (x  -  y) 


TO  FACTOR  A  TRINOMIAL  OF  THE  FORM  X2  +  CtX  +  b.     99 

65.  To  Factor  a  Trinomial  of  the  Form  x2+ax-\-6. — 

Let  x'2  +  ax  +  b  be  any  trinomial  in  which  the  coefficient 
of  a;2  is  +1 ,  and  the  signs  of  a  and  b  either  pins  or  minus. 

Before  proceeding  to  explain  this  case  of  resolution  into 
factors,  the  student  is  advised  to  refer  to  Art.  40,  and 
examine  the  relation  that  exists  between  two  binomial 
factors  and  their  product.  Attention  was  there  called  to 
the  way  in  which,  in  forming  the  product  of  two  binomials, 
the  coefficients  of  the  different  terms  combined  so  as  to  give 
a  trinomial  result. 

Therefore,  in  the  converse  problem,  namely,  the  resolution 
of  a  trinomial  expression  into  its  component  binomial  factors, 
we  see,  by  reversing  the  results  of  Art.  40,  that  any  trino- 
mial may  be  resolved  into  two  binomial  factors,  when  the 
first  term  is  a  square,  and  the  coefficient  of  the  secoud  term 
is  the  sum  of  two  quantities  whose  product  is  the  third  term. 
Hence  the  following 

Rule. 

The  first  term  of  each  factor  is  x,  and  the  second  terms 
are  two  numbers  whose  Algebraic  sum  is  the  coefficient  of  the 
second  term,  and  whose  product  is  the  third  term. 

The  application  of  this  rule  will  be  easily  understood  from 
the  following 

EXAMPLES. 

1.  Resolve  into  factors  x"  +  llx  +  24. 

Here  the  first  term  of  each  binomial  factor  is  x,  and  the 
second  terms  of  the  two  binomial  factors  must  be  two 
numbers  whose  sum  is  11  and  whose  product  is  24.  It  is 
clear  therefore  that  they  must  be  +8  and  +3,  since  these 
are  the  only  two  numbers  whose  sum  is  11  and  whose  product 
is  24. 

.-.     x2  +  11a  +  24  =  (x  +  8)(<e  +  3). 

2.  Resolve  into  factors  x2  —  7x  +  12. 

The  first  term  of  each  factor  is  x,  and  the  second  terms 
of  the  factors  must  be  such  that  their  sum  is  —  7,  and  their 


100  EXAMPLES. 

product  is  +12.     Hence  they  must  both  be  negative,  and  it 
is  easy  to  see  that  they  must  be  —4  and  —3. 

.-.     x2  -  7x  +  12  =  (x  -  4) (as  -  3). 

3.  Resolve  into  factors  x2  +  5x  —  24. 

The  first  term  of  each  factor  is  x,  and  the  second  terms 
of  the  factors  must  be  such  that  their  Algebraic  sum  is  -f-5, 
and  their  product  is  —24.  Hence  they  must  have  opposite 
signs,  and  the  greater  of  them  must  be  positive  in  order  to 
give  the  positive  sign  to  their  sum.  It  is  easy  to  see  there- 
fore that  they  must  be  +8  and  —3. 

...     a;2  +  5se  -  24  =  (x  +  8){x  -  3). 

4.  Resolve  into  factors  x2  —  x  —  56. 

The  first  term  of  each  factor  is  as,  and  the  second  terms 
of  the  factors  must  be  such  that  their  Algebraic  sum  is  —  1 , 
and  their  product  is  —56.  Hence  they  must  have  opposite 
signs,  and  the  greater  of  them  must  be  negative  in  order  to 
give  its  sign  to  their  sum.  The  required  terms  are  therefore 
-8  and  +  7. 

.-.     x2  -  x  -  56  =  {x  -  8)(as  +  7). 

Note.  —  In  examples  of  this  kind  the  student  should  always  verify 
his  results,  by  forming  the  product,  mentally,  of  the  factors  he  has 
chosen,  as  in  Art.  40. 

Resolve  into  factors 


5. 

a2  +  3a  +  2. 

Ans.  (a  +  l)(a  +  2) 

6. 

x2  -  Ha;  +  30. 

(,•-  6)(.r-5) 

7. 

a'2  -  7a  +  12. 

(a  -  4)  (a  -  3) 

8. 

x2  —  15.r  +  56. 

(x  -  8)(x  -  7) 

9. 

x2  -  19a  +  90. 

(x  -  9)  (as  -  10) 

10. 

x2  +  x  -  2. 

(x  +  2)(x~  1) 

11. 

x2  +  x  -  6. 

(.*  +  3)(*-  2), 

12. 

x2  -  2x  -  3. 

O-  3)(.r  +  1) 

13. 

x2  +  2x  -  3. 

(x  +  3)(x-  1) 

14. 

x2  +  x  —  56. 

(»  +  8)  (x  -  7) 

15. 

x-  +  3x  —  40. 

(a,-  +  8)(a--5) 

TO  FACTOR  A  TRINOMIAL  OF  THE  FORM  ax2  +  bx  +  c.    101 

1G.  x2  -  4x  -  12.  Arts,  (x  —  6)  (a;  +  2). 

17.  a;2  -  a;  -  20.  (a;  -  5)  (a;  +  4). 

18.  a2  -  4a  -  21.  (a  -  7) (a  +  3). 

19.  a2  +  a  —  20.  (a  +  5)  (a  —  4). 

20.  a2  -  4a  -  117.  (a  -  13)  (a  +  9). 

21.  a;2  +  9a;  -  3G.  (a;  +  12)(aj  -  3). 

Note.  —  If  the  term  containing  x1  is  negative,  enclose  the  whole 
expression  in  a  parenthesis  with  the  minus  sign  prefixed.  Then  factor 
the  expression  within  the  parenthesis  as  in  the  preceding  examples, 
and  change  the  signs  of  all  the  terms  of  one  of  the  factors.     Thus 

22.  Factor  90  +  9x  —  x2. 
90+9x-x2=-(x2-9x-90)  =  -(x-15)(x+6)  =  (lo-x)(x+6). 

23.  Factor  240  +  x  -  x2.  Arts.  (16  —  a?)  (a;  +  15). 
21}  Factor  85  +  12a;  -  a;2.  (17  —  x)  (x  +  5). 
g;  Factor  110  -  x  -  x2.  (10  -  x)  (x  +  11). 
2G.  Factor  152  +  11a;  -  x2.  (19  -  x)(x  +  8). 

66.  To  Factor  a  Trinomial  when  the  Coefficient 
of   the    Highest    Power   is   not   Unity.  —  Let  it  be 

required  to  factor  3a;2  +  14a:  -f-  8„ 

The  first  term  3a;2  is  the  product  of  3a;  and  x. 

The  third  term  8  is  the  product  of  2  and  4  or  1  and  8. 

The  middle  term  14a;  is  the  result  of  adding  together  the 
two  products  3a;  and  2  and  x  and  4,  or  3a;  and  4  and  a;  and  2, 
or  3a;  and  1  and  x  and  8,  or  3a;  and  8  and  x  and  1. 

Taking  the  first  products  we  have  3a;  x  2  +  x  x  4  =  10a;; 
this  combination  therefore  fails  to  give  the  correct  middle 
term. 

Next  try  the  second  products,  and  get  3a;x4  +  a;x  2  =  14a;, 
which  is  the  correct  value  of  the  middle  term. 

.-.     3a;2  +  14a;  +  8  =  (3a;  +  2)  (a;  +  4). 

The  beginner  will  frequently  find  that  it  is  not  easy  to 
select  the  proper  factors  at  the  first  trial.  Practice  alon* 
will  enable  him  to  detect  at  a  glance  whether  any  two  factors 
are  the  proper  ones. 


102  EXAMPLES. 


EXAMPLES. 


1.  Resolve  into  factors  l\n^  -f-  29a;  —  15. 

Write  down  (7a;  5)  (2a;  3)  foi  a  first  trial,  noticing  that 
3  and  5  must  have  opposite  signs.  These  factors  give 
14a;2  and  —15  for  the  first  and  third  terms.  But  since 
7x3—  2x5  =  11,  this  combination  fails  to  give  the 
correct  coefficient  of  the  middle  term. 

Next  try  (7a;  3)  (2a;  5). 

Since  7x5  —  3x2  =  29,  these  factors  will  be  correct 
if  we  insert  the  signs  so  that  the  positive  will  predominate. 
.-.     14a;2  +  29a;  -  15  =  (7a;  -  3)  (2a-  +  5). 

(Verify  by  mental  multiplication). 

It  is  not  usually  necessary  to  put  down  all  thes^  steps. 
After  a  little  practice  the  student  will  be  able  to  examine  the 
different  cases  rapidly,  and  to  reject  the  unsuitable  combina- 
tions at  once. 

In  the  factoring  of  such  expressions  as  these  the  following 
hints  are  very  useful : 

(1)  If  the  third  term  of  the  trinomial  is  positive,  then  the 
second  terms  of  its  factors  have  both  the  same  sign  as  the 
middle  term  of  the  trinomial. 

(2)  If  the  third  term  of  the  trinomial  is  negative,  then 
the  second  terms  of  its  factors  have  opposite  signs. 

2.  Resolve  into  factors  5ar  +  17a;  +  6 (1) 

5a-2  -  17a;  +  G (2) 

5a;2  +  13a;  -6 (3) 

5a;2  -  13a-  -  6 f4) 

In  (1)  we  notice  that  the  factors  which  give  G  are  botb 
positive. 

In  (2)  we  notic"1  that  the  factors  which  give  G  are  both 
negative. 

In  (3)  we  ....ace  that  the  factors  which  give  G  have 
opposite  signs. 

In  (1)  we  notice  that  the  factors  which  give  G  have 
opposite  signs. 


EXAMPLES.  103 

Therefore  for  (1)  we  write  (5a;  +    )(#+). 

(2)  we  write  (5a:.—    )  (x  —   ). 

(3)  we  write  (5a;       2)  (a;      3),  noticing  that 

2  and  3  have  opposite  signs. 

(4)  we  write  (5a;       2)  (a;      3),  noticing  that 

2  and  3  have  opposite  signs. 
Since  5  x  3  +  1  x  2  =  17,  we  see  that 

5a;2  +  17a;  +  6  =  (5»  +  2)  (x  +  3). 
5a;2  —  17a;  +  6  =  (5x  —  2)  (as  —  3). 
And,  since  5x3  —  2  x    1   =    13,  we  have  only  to  insert 
the  proper  signs  in  each  factor. 

In  (3)  the  positive  sign  must  predominate. 
In  (4)  the  negative  sign  must  predominate. 
Therefore  5a;2  +  13a;  -  6  =  (5a;  —  2)  (a;  +  3). 
5a;2  -  13a;  -  6  =  (5a;  +  2)  (a;  -  3). 
More  generally,  trinomials  of  the  form  aa;2  +  bx  +  c,  (a 
not  a  square)    may  be  factored  more  readily  as  follows : 
Multiplying  by  a  we  get  a2x2  +  bax  +  ac.     Writing  z  for 
ax,  this  becomes  z2  +  bz  +  ac.      Factor  this  trinomial  by 
Art.  G5,  replace  the  value  of  z,  aud  divide  the  result  by  a. 
Thus, 

3.  Resolve  into  factors  6a;2  —  13a;  +  6. 
Multiplying  by  6  we  get  (6a;)2  -  13(6a-)  +  36. 

Putting  z  for  6a;  we  get 

z2  -  13s  +  36, 
which,  being  factored,  gives 

(*-  9)(.-4). 
Hence  the  required  factors  of  Gx2  —  13a:  +  6  are 

£(6a;  -  9)  (6a;  -  4)  =  (2a;  -  3)  (3a:  -  2). 
Resolve  into  factors 

4.  3a;2  +  5a;  +  2.  Ans.  (3x  +  2)  (a;  +  1). 

5.  2x2  +  5x  +  2.  (2b  +  l)(a!  +  2). 

6.  2a;2  4-  7x  +  6.  (2a;  +  3)  (a;  +  2). 

7.  3a;2  +  8x  +  4.  (3a;  +  2)  (a:  +  2). 

8.  2a;2  +  llx  -f-  5.  (2a;  +  l)(as  +  5). 


104     TO   FACTOR   THE   DIFFERENCE   OF   T\V(    SQUARES. 


9.  3a;2  +  10a  +  3.  Ans.  (3a  +  l)(a  +  3) 

:j.  3ar  +  11a  +  6.  (3a  +  2)  (a;  +  3) 

If.  4a2  +  llic  -  3.  (4a  -  l)(a  +  3) 

12.  3a;2  +  x  -  2.  (3a?  -  2)  (a:  +  1) 

i:;.  2a;2  +  3a;  -  2.  (2a;  -  l)(a  +  2) 

14.  2a;2  +  15a;  -  8.  (2a;  -  l)(a  +  8) 

15.  3a2  -  19a;  -  14.  (3a;  +  2)  (a-  -  7) 
1G.  G.«2  -  31a;  +  35.                             (3a;  —  5)  (2a;  —  7) 

17.  3a?  +  19a  -  14.  (3a;  -  2)  (a  +  7) 

18.  4a;2  +  a  -  14.  (4a  -  7)  (a  +  2) 

67.  To  Factor  an  Expression  which  is  the  Differ- 
ence of  Two  Squares.  —  From  (3)  of  Art.  41,  we  see 
that  the  difference  of  two  squares  is  equal  to  the  product  of 
the  sum  and  difference  of  their  square  roots.  Therefore, 
conversely,  to  find  the  factors  we  have  the  following 

Rule. 
Extract  the  square  roots  of  the  two  terms;  take  the  sum  of 
the  results  for  one  factor,  and  the  difference  for  the  other. 

EXAMPLES. 

1.  Resolve  into  factors  25a;2  —  16?/2. 

25a2  -  16y»  =  (5a)2  -  (4y)2. 
Therefore  the  first  factor  is  the  sum  of  5a;  and  4y,  and  the 
second  is  the  difference  of  5a;  and  ly. 

.-.     25a;2  -  lGy2  =  (5a  +  4y)(5a  -  4y). 
The  intermediate  steps  may  usually  be  omitted. 
Resolve  into  factors 

2.  x-  —  9a2.  Ans.  (x  +  3a)  (x  -  3a). 

3.  121  -  x2.  (11  +  »)(11  -  a). 
A.  f  -  25a3.  (y  +  5a) (y  -  5a). 
5.  36a2  -  2563.  (6a  +  56) (6a  -  56). 
G.  xhf  -  36.  (xy  +  6)  (ay  -  6). 
7.   a-lr  -  -UihlK                           (ab  +  Sod)  (aft  -  2cd). 


TO  FACTOR    THE   DIFFERENCE    OF   TWO   SQUARES.     105 


8.  81a4  -  49a4 

P  9a4  -  121. 

|  lb),  a6  -  25. 

(Tp.  a4a2  -  49. 

12.  a2  -  G4a6. 


J.ns.  (9a2  +  7a2)  (9a2  -  7a2). 

(3a2  +  11) (3a2  -  11). 

(xs  +  5)  (a;3-  5). 

(a2a  +  7)(a2a  -  7). 

(a  +  8a3)  (a  -  8a3) . 


68.  When  One  or  Both  of  the  Squares  is  a 
Compound  Expression.  —  Here  the  same  method  is 
employed  as  in  the  last  Article. 


EXAMPLES. 

\J  Resolve  into  factors  (2a  +  b)2  —  9a2. 
he  sum  pf  2a  +  b  and  3a  is  2a  +  b  +  3a, 
and  their  difference  is  2a  +  b  —  3a. 

.-.      (2a  +  b)2  -  9a2  =  (2a  +  b  +  3a)  (2a  +  6  -  3a). 
If  the  factors  contain  like  terms  they  should  be  collected 
so  as  to  give  the  result  in  its  simplest  form. 

X2).   Resolve  into  factors  (5a  +  8y)2  —  (4a  —  3^/)2. 
/  The  sum  of  5a  +  8y  and  4a  —  3?/ 
is  5a  +  8y  +  4a  —  3?/  =  9a  +    5y, 

and  their  difference  is 

5a  -f  8y  —  4a  +  oy  =    a  +  lly. 
.-.      (5a  +  8y)2  -  (4a  -  3y)2  =  (9a  +  5y)  (a  +  lly). 
Resolve  into  factors 

Ans.  (a  +  6  +  c)  (a  +  &  —  c) 

4.  (a  -  b)°~  -  c2.  (a  -  b  +  c)  (a  -  b  -  c) 

5.  (a  +  y)"  -  4z2.  (a  +  y  +  2z)  (a  +  y  -  2s) 
(a  +  2y  +  a)  (a  ■+-  2y  —  a) 
(a  -  2a  +  b)  (a  -  2a  -  b) 

(2a  -  3a  +  3c)  (2a  —  3a  —  3c) 

(2a  +  y)y 

y{-2x  -  y) 

(a  +  5y)  (a  +  y) 

4)2.  (12a-  l)(2a  +  7) 


(a  +  6)2  -  c2. 
(a  -  &)2  -  c2. 
,.    (^  +  yy  _  4z2. 
(a  +  2t/)2  -  a2, 
(a  -  2a)2  -  b2. 
(2a  -  3a)2  -  9c2. 
(a  +  y)'2  -  a2. 
x2  -  (y  -  a)2, 
(a  +  By)2  -  4?/2. 
(7a  +  3)2  -  (5a 


106     TO  FACTOR   T1IK    DIFFERENCE   OF    TWO   SQUARES. 

69.  Compound  Quantities  expressed  as  the  Dif- 
ference of  Two  Squares.  —  Compound  expressions,  by 
suitably  grouping  the  terms,  can  often  be  expressed  as  the 
difference  of  two  squares,  and  so  be  resolved  into  factors. 

EXAMPLES. 

1.  Resolve  into  factors  a,    +  2ax  +  a?2  —  4b'2. 
From  (1)  of  Art.  41,  a2  +  2ax  +  x1  =  (a  +  x)'2. 

.-.     a2  +  2ax  +  x2  -  4&2  =  (a  +  a)2  -  462, 
which  by  Art.  G8  =  (a  +  *  +  26)  (a  +  •*'  -  26). 

2.  Resolve  into  factors  9a'2  —  c2  -f  Acx  —  4a:2. 
9a2  —  c2  +  4ca;  —  4a;'2  =  9a2  —  (c2  —  4«c  +  4a;2) 

=  9a2  -  (c  -  2a;)2  by  (2)  of  Art.  41, 
=  (3a  +  c  -  2a)  (3a  -  c  +  2a;). 

3.  Resolve  into  factors  24xy  +  25  —  lGar  —  9y2. 
24xy  +  25  -  16a;2  -  9y'2  =  25  -  (IGa;2  -  24a^  +  9y2) 

=  25  -  (4a;  -  3?/)2by  (2)  of  Art.  41, 
=  (5  +  4aj-3y)(5-4a>  +  3y). 

4.  Resolve  into  factors  26d  —  c2  —  a2  +  d2  +  62  +  2ac 
Here  we   see   that   the   expression   is   composed   of   two 

trinomials,  each  of  which  is  the  square  of  a  binomial  [(1) 
and  (2)  of  Art.  41]. 

.-.     2bd  —  c2— a2+d2+b2  +  2ac  =  b2  +  2bd  +  d2-(a2-2a<-  +  r-) 

=  (6+<Z)s-(a-c)2 

=  (P+d+a-c)(b+d-a+c). 

Resolve  into  factors 

5.    a;2  +  2ayy  +  y2  —  a2.        Ans.  (x  -f-  y  +  a)  (x  +  y  —  a). 

G.    x2  -  Gax  +  9«2  -  1 6b2.       (x  -  3a  4-46)  {x  -  3a  -  46)  . 

7.  4a2  +  4ab  +  62  -  9c2.       (2a  +  b  +  3c)  (2a  4-  6  -  8c). 

8.  x2  +  a2  4-  2aaj  -  ya.         (a-  +  a  +  y)  (x  +  a  -  y). 
0.   ca  _  ^  -  t/2  +  2<By.  (c  4-  a-  -  y)  (c  -  a?  +  y) . 

10.    as*  +  y2  +  2  ay  -  4afya .       (x  +y+  2xy)  (x  +y-  2xy) . 


MISCELLANEOUS   CASES   OF  FACTORING.  107 

70.  To  Factor  an  Expression  which  can  be 
Written  as  the  Sum  or  the  Difference  of  Two 
Cubes.  —  Such  expressions  as  these  may  be  resolved  into 
factors  by  Art.  51. 

EXAMPLES. 

1.  Resolve  into  factors  8a;3  —  27/. 

By  I.  of  Art.  51,  this  is  divisible  by  2x  —  Sy. 
.-.     8x3  -  27/  =  (2a;)3  -  (Sy)3 

=  (2x  -  3?/)  (4a;2  +  Gxy  +  9/). 
Note.  —  The  middle  term  6xij  is  the  product  of  2x  and  oy. 

2.  Resolve  into  factors  343a;6  +  27 ys. 

343a;6  +  27ys  =  (7a;2  +  3y)  (49a;4  -  21x2y  +  9if), 
(by  III.  of  Art.  51). 

Resolve  into  factors 

3.  8a3  -  27/.  Ans.  (2a  -  3?/) (4a2  -f-  Gay  +  9/). 

4.  1  -  343b8.  (1  -  7a)  (1  +  7x  +  49a;2). 

5.  a3b3  +  512.  (ab  +  8)  (a362  -  Sab  +  64). 

6.  343  +  8a;3.  (7  +  2x)  (49  -  14a;  +  4a;2). 

7.  216a;3  -  343.  (6x  -  7)  (36a;2  +  42a;  +  49). 

8.  27a;3  -  64/.  (3a;  -  Ay)  (9a;2  +  \2xy  -+-  16/). 

9.  64a;6  +  125/.  (4a;2  -f-  5/)  (16a;4  -  20x2y  +  25/). 

10.  216a;6  -  b3.  (Gx2  -  b)  (36a;4  +  6a;2&  +  b2) . 

11.  a3  +  34363.  (a  +  76)  (a2  -  7a6  +  4962). 

71.  Miscellaneous  Cases  of  Resolution  into  Fac- 
tors.—  When  an  expression  can  be  arranged  as  the  differ- 
ence of  two  squares,  it  may  be  factored  either  by  (II.)  of 
Art.  51  or  by  (3)  of  Art.  41 .  It  will  be  found  the  simplest, 
however,  first  to  factor  by  the  second  method,  using  the  rule 
for  factoring  the  difference  of  two  squares  (Art.  67). 


108  EXAMPLES. 

EXAMPLES. 

1.  Resolve  into  factors  16a4  —  8164. 

16a4  -  81b4  =  (4a2  +  962)  (4a2  -  962)  (Art.  67) 

=  (4a2  +  962)(2a  +  36)  (2a  -  3b)  (Art.  67). 

2.  Resolve  into  factors  x6  —  y*. 

xa  -  y6  =  (xs  +  ys)(x3  -  y3)  (Art.  67) 

=  (x  +  y)  (a2  -  ®y  +  r)  (*  -  y)  (a2  +  xy  +  f) 

(Art.  51,  I.  and  III.). 

The  student  should  be  careful  in  every  case  to  remove  all 
monomial  factors  that  are  common  to  each  term  of  an  ex- 
pression, and  place  them  outside  a  parenthesis,  as  explained 
in  Art.  63. 

3.  Resolve  into  factors  28a;4?/  +  Wy  -  60a;2?/. 
28k4?/  +  64a;3?/  -  60.t2?/  =  4.t-2?/(7a;2  +  16a;  -  15) 

=  Ax~y(lx  —  5)(aj  +  3)  (Art.  66). 

4.  Resolve  into  factors  a;3a2  —  8y3a2  —  4a;3&2  +  32ysb2. 
xW-  Sfa2-  4x862+  32y*b*=  a2(.r3-  8f)  -  4b2(x3-  8y3) 

=  (a«-8y8)(a2-46a) 

=  {x-2y)(x2+  2xy+ ±y2)(a+ 2b){a—2b). 

5.  Resolve  into  factors  4a;2  —  25?/2  +  2a;  +  5?/. 

4a;2  —  2by*  +  2a;  +  by  —  (2a;  +  5?y)  (2a;  —  by)  +  2a;  +  5?/ 
=  (2a;  +  5//)  (2a;-  by  +  1). 
Resolve  into  two  or  more  factors 

6.  a2  —  y-  —  2yz  —  z'2.         Ans.  (a  +  y  +  z)(a  —  y  —  z) . 

7.  6x-2  -  a?  -  77.  (3a;  -  11)  (2a:  +  7). 

8.  a;6-4096.      (a?+4)  (aj2-4a?+16)  (a?-4)  (a;2+4a;+16). 

9.  x2  —  a2  +  y2  —  2xy.         (x  —  y  +  a)(x  —  y  —  a). 

10.  aca;2  —  bcx  +  adx  —  bd.  (ex  +  d)(aai  —  6). 

11.  (a  +  6  +  c)2  -  (a  -  &  -  c)a.  4a(6  +  c). 
Other  expressions  which,  by  a  slight  modification,  can  be 

arranged  as  the  difference  of  two  squares,  may  be  factored 
by  Art.  67. 


EXAMPLES.  109 

'12.    Resolve  into  factors  x4  +  x2y2  +  y*. 
xi  _|_  x-y'2  +  ?/4  =  x4  +  2.r'7/2  +  ?/4  —  ar?/2 

=  (•'■'"  +  y-)'1  -  »V 

=  (a2  +  2T  +  xy)  (x2  +  2/2  -  xy)  ■ 

13.  Resolve  into  factors  x*  —  loo;2//2  +  Oy4. 
x*  _  I5a;y  +  dy*  =  (x2  -  Sy2)2  -  dx2y2 

=  (<c2  —  3?/2  +  3x?/)  (a2  -  Sy2  -  %xy) . 

Expressions  which  can  be  put  into  the  form  xz  ±  —  may 

be  factored  by  the  rules  for  resolving  the  sum  or  the  differ- 
ence of  two  cubes  (Art.  70) . 

14.  Resolve  into  factors  —  —  27y6. 

x3 

■  J-27**=(Cf-^>* 

15.  Resolve  a2x3  -  —  -  a;3  +  4  iuto  four  factors. 

f  y 

oW-^-a»+  i  =  *»(«S-l)  --^(«2-l) 

?/3  yz  y 


=  («2-l) 


W 


=(a+1)(„_1}HX^+!+i). 


16.  Resolve  a9  —  64a3  —  a6  +  64  into  six  factors. 
The  expression 

=  a3(a6-64)-(o,6-64) 
=  (a6-64)(«3-l) 
=  (a8+8)(a8-8)(a8-l) 

=  (a  +  2)  (a2-2a+4)  (a-2)  (a2+2a+4)  (a-1)  («2  +  a  +  1 ) . 
Resolve  into  factors 

17.  a4  +  16a;2 +  256.         ^,JS.  (asF+4as+16)  (a*-4as+16). 

18.  s4  +  ?/4  -  7xV.         (a,-2  +  3a^  +  ?/)  (a;2  -  3zy  +  */2) . 

19.  81a*+9a2&2+64.         (9a2+ 3a&  +  62)  (9a2  -  3a&  +  &2) . 

20.  a;4-  19x22/2 +25/.         (x2+  3ay-  5y2)  (a2-  3xy-  by2) . 


110  EXAMPLES. 

By  a  skilful  use  of  factors,  the  actual  processes  of  multi- 
plication and  division  can  often  be  partially  or  wholly 
avoided. 

21.  Multiply  2a  +  36  —  c  by  2a  —  36  +  c. 

The  product  =  [2a  +  (36  -  c)][2a  -  (36  -  c)] 
=  (2a)2  -  (36  -  c)2  [(3)  of  Art.  41] 

=  4a2  -  9b2  +  66c  -  c2. 

22.  Divide  the  product  of  x2  —  bxy  +  Gy2  and  x  —  4y 
by  x2  —  Ixy  -+-  I2y2. 

We  might  multiply  the  first  two  expressions  together  and 
then  divide  the  result  by  the  third.  But  by  factoring  the 
first  and  third  expressions,  and  denoting  the  division  by 
means  of  a  fraction  (see  Art.  78),  the  work  will  be  much 
shorter. 

Thus,  the  required  quotient 

_  (x2  —  oxy  +  Gy2)  (x  —  4y) 

x2  —  Ixy  +  I2y2 
_  (x  —  3y)  (x  —  2?/)  (x  —  4y) 

(x  -  Ay)  (x  -  3y) 
=  x  -  2y. 

23.  Divide  the  product  of  2x2  +  x  —  6  and  Gx2  -  5x  -f  1 
by  3x2  +  bx  —  2.  Ans.  (2x  -  3)(2as  -  1). 

Find  the  product  of 

24.  2x—  7y  +  3z  and  2x+7y— 3z.       4x2—49y2+42yz—9z2. 

25.  x3+2x2y+2xy2+y3  and  Xs  —  2x2y  +  2xy2— y3.        x6—y6. 

26.  xs  -4x2  +  8x  -  8  and  x3  +  4a;2  +  8a;  +  8.         xe  -  64. 
Divide 

27.  2a,-(a;2- 1 )  (x  -f  2)  by  x2  +  x-2.  2x(x  +  1 ) . 

28.  (x2  +  7x  +  10)  (x  +  3)  by  x2  +  5x  +  6.         x  +  5. 

29.  5a?(as- 11)  (a;2-a;- 156)  by  .rs+a:2-132a;.     5(«-18). 

30.  a,9  -  69  by  (a2  +  ab  +  62)  (a,6  +  a3bs  +  66) .         a  -  6. 

31.  [a3  +  (a  -  6)oj  -  aft]  |V  -  (a  -  &)a  -  06]  by 
a3  +  (a  +  6).«  +  ab.  Ans.  (as  —  a)(x  -  b). 


GREATEST   COMMON  DIVISOR  —  DEFINITIONS.        Ill 

GREATEST    COMMON    DIVISOR. 

72.  Definitions.  —  A  Common  Divisor  of  two  or  more 
expressions  is  an  expression  that  will  divide  each  of  them 
exactly. 

Hence,  every  factor  common  to  two  or  more  expressions  is 
a  common  divisor  of  those  expressions  (Art.  62) . 

Thus,  in  4a26,  6a3b2,  and  a4b3,  a2  occurs  as  a  factor  of  each 
quantity  ;  b  also  occurs  as  a  factor  of  each  quantity  ;  a2  and 
b  are  therefore  common  divisors  of  these  three  quantities. 

The  Greatest  Common  Divisor  of  two  or  more  Algebraic 
expressions  is  the  expression  of  highest  degree  (Art.  18) 
which  will  divide  each  of  thern  exactly. 

Note.  —  The  term  greatest  common  divisor,  which  has  been  adopted 
from  Arithmetic,  does  not  imply  in  Algebra  that  it  is  numerically  the 
greatest,  but  that  it  is  the  factor  of  greatest  degree.  The  student  is 
cautioned  against  being  misled  by  the  analogy  between  the  Algebraic 
and  the  Arithmetic  greatest  common  divisor.  He  should  notice  that 
no  mention  is  made  of  numerical  magnitude  in  the  definition  of  the 
Algebraic  greatest  common  divisor.  In  Arithmetic,  the  greatest 
common  divisor  of  two  or  more  whole  numbers  is  the  greatest  whole 
number  which  will  exactly  divide  each  of  them.  But  in  Algebra,  the 
terms  greater  and  less  are  seldom  applicable  to  those  expressions  in 
which  definite  numerical  values  have  not  been  assigned  to  the  various 
letters  which  occur.  Besides,  it  is  not  always  true  that  the  Arith- 
metic greatest  common  divisor  of  the  values  of  two  given  expressions 
obtained  by  assigning  any  particular  values  to  the  letters  of  those 
expressions,  is  the  numerical  value  of  the  Algebraic  greatest  common 
divisor  when  those  same  values  of  the  letters  are  substituted  therein, 
as  will  be  shown  later  (Art.  74).  For  this  reason,  some  writers  have 
used  the  terms,  highest  common  divisor,  and  highest  common  factor, 
instead  of  the  term  greatest  common  divisor.  But  to  avoid  employing 
a  new  phrase,  and  in  conformity  with  well-established  usage,  we  shall 
retain  the  old  term  greatest  common  divisor. 

The  abbreviation  G.  C.  D.  will  often  be  used  for  shortness  instead 
of  the  words  greatest  common  divisor. 

73.  The  Greatest  Common  Divisor  of  Monomials, 
and  of  Polynomials  which  can  be  easily  Factored. 

—  Let  it  be  required  to  find  the  greatest  common  divisor  of 
21a4z3y,  35a2afy,  28asxY,  and  Ua5x2y\ 


112       GREATEST  COMMON  DIVISOR   OF  MONOMIALS. 

By  separating  each  expression  into  its  prime  factors,  we 
have  2la*x3y    =  7  X  3aaaaxxxy. 

35a2x4y    =  7  X  baaxxxxy. 
28a3x2yi  =  7  x  2  x  2aaaxxyyyy. 
14abx2y2  =  7  X  2aaaaaxxyy. 
By  examining  these  expressions  we  find  that  7,  aa,  xx, 
and  y  are  the  only  factors  common  to  all  of  them.     Hence 
all  the  expressions  can  be  exactly  divided  by  either  of  these 
factors,  or  by  their  product,  la2x2y,  which  is  therefore  their 
greatest  common  divisor. 

Find  the  G.  C.  D.  of  4cxs  and  2cx3  +  4cV. 
Resolving  each  expression  into  its  factors,  we  have 
Acx3  —  2cx2  x  2x. 
2cx3  +  Ac2x2  =  2cx2{x  +  2c). 
Here  it  is  clear  that  both  expressions  are  divisible  (1)  by 
2,  which  is  the  numerical  greatest  common  divisor  of   the 
coefficients,  (2)  by  c,  and  (3)  by  x2. 

.-.     G.  C.  D.  =  2cx\ 
Find  the  G-.  C.  D.  of  3a2  +  9a&,  a3  —  dab2,  a3  +  6a2&  +  9a&2. 
Resolving  each  expression  into  its  factors,  we  have 
3a2  +  9a6  =  3a  (a  +  36). 
a3  -  9a62  =  a(a  +  3b)  (a  -  3b). 
a3  +  Ga2b  +  dab2  =  a(a  +  3b)  (a  +  3b). 
.-.     G.  C.  D.  =  a(a  +  36). 
Find  the  G.  C.  D.  of  x(a  —  a)2,  ax  (a  —  x)3,  2ax(a  -  x)\ 
Resolving  into  factors,  we  have 
x(a  —  x)2  =  x(a  —  x)(a  —  x). 
ax(a  —  x)s  =  aoj(a  —  x)(a  —  x)(a  —  x). 
2ax(a  —  a-)4  ==  2a,x(a  —  x)(a  —  x){a  —  a;)  (a  —  x). 
.-.     G.  C.  D.  =  x{a  -  x)2. 
Hence  the  following 

Rule. 

Resolve  each  expression  into  its  prime  factors,  and  take  the 
product  of  all  the  factors  common  to  all  the  expressions, 
giving  to  each  factor  the  highest  power  tohich  is  common  to  cdl 
the  given  expressions. 


GREATEST   COMMON  DIVISOR   OF  POLYNOMIALS.     113 

EXAMPLES. 
Find  the  G.  C.  D.  of 

1.  4a&2,  2a26,  6ab3.  .  Am.  2ab. 

2.  3xY,  x3y2,  x2y3.  x2y2. 

3.  6xy2z,  8x2y3z2,  4xyz2.  2xyz. 

4.  5asb3,  loabc2,  Wa2b2c.  bob. 

5.  9x2y2z2,  12xyzz,  6x3y2z3.  Sxy2z. 

6.  8a2x,  6abxy,  10abx3y2.  2ax. 

7.  a2  +  ab,  a2  -  b2.  a  +  6. 

8.  (x  +  y)2,  x2  -  y2.  x  +  y. 

9.  x3  +  x2y.  x3  -\-  y3.  x  +  y. 

10.  a3  —  a2x,  a3  —  ax2,  a4  —  ax3.  a(a  —  x). 

11.  x4  -  27a3x,  (x  -  3a)2.  x  -  3a. 

12.  xy  -  y,  »V  -  xy.  y(x  -  1). 

13.  ax2  +  2a2x  +  a3,  2ax2  -  4a2x  -  6a3,  3 (ax  +  a2)2. 

-4ws.  a(x  +  a). 

74.  The  Greatest  Common  Divisor  of  Expressions 
that  cannot  be  Readily  Resolved  into  Factors. — 
To  find  the  G.  C.  D.  in  such  cases,  we  adopt  a  method 
analogous  to  that  used  in  Arithmetic  for  finding  the  G.  C.  D. 
of  two  or  more  numbers. 

The  method  depends  on  two  principles. 

1.  If  an  expression  contain  a  certain  factor,  any  multiple 
of  that  expression  is  divisible  by  that  factor. 

Thus,  if  F  divides  A  it  will  also  divide  mA.  For  let  a 
denote  the  quotient  when  A  is  divided  by  F;  then  A  =  aF; 
therefore  mA  =  maF\  and  therefore  F  divides  mA. 

2.  If  two  expressions  have  a  common  factor,  it  will  divide 
their  sum  and  their  difference;  and  also  the  sum  and  the 
difference  of  any  midtiple  of  them. 

Thus,  if  F  divides  A  and  B,  it  will  divide  mA  ±  nB.  For 
since  F  divides  A  and  B,  we  may  suppose  A=aF,  and  B—bF; 
therefore  mA  ±  nB  =  maF  ±  nbF 

=  F(ma  ±  nb). 
Therefore  F  divides  mA  ±  nB, 


114     GREATEST   COMMON   DIVISOR   OF  POLYNOMIALS. 

We  can  now  prove  the  rule 'for  finding  the  G.  C.  D.  of  any 
two  compound  Algebraic  expressions. 

Let  A  and  B  denote  the  two  expressions.  Let  them  be 
arranged  in  ascending  or  descending  powers  of  some  common 
letter  ;  and  let  the  highest  power  of  that  letter  in  B  be  either 
equal  to  or  greater  than  the  highest  power  in  A. 

Divide  B  by  A  ;  let  p  be  the  quotient  and  C  the  remainder. 
Suppose  C  to  have  a  simple  factor  m.  Remove  this  factor, 
and  so  obtain  a  new  divisor  D.  Suppose  further,  that  in 
order  to  make  A  divisible  by  D  it  is  necessary  to  multiply  A 
by  a  simple  factor  n.  Divide  nA  by  D  ;  let  q  be  the  next 
quotient,  and  E  the  remainder.  Divide  D  by  E ;  let  r  be 
the  quotient,  and  suppose  that  there  is  no  remainder.  Then 
E  will  be  the  G.  C.  D.  required. 

The  operation  of  division  will  stand  thus : 

A)B(p 
pA 

mjc 

D)nA(q 
qD 

E)D(r 
rE 

First,  to  show  that  E  is  a  common  divisor  of  A  and  B. 
From  the  above  division  we  have  the  following  results : 

D  =  rE. 
nA  =  qD  +  E  =  qrE  +  E      =  {qr  +  \)E. 

B=  pA  +  C  =  pA   +  mD  =  1WE  +  pE  +  mrE 


fpqr  -f  p  \ 


Therefore  E  is  a  common  divisor  of  A  and  B. 

Second,  to  show  that  E  is  the  greatest  common  divisor  of 
A  and  B. 


EXAMPLES.  115 

By  (2)  of  this  Art.  every  common  factor  of  A  and  B 
divides  also  B  —  pA,  that  is  O,  and  therefore  D  (since  m  is 
a  simple  factor) .  Similarly  as  it  divides  A  and  D  it  divides 
nA  —  qD,  that  is  E.  But  no  expression  of  higher  degree 
than  E  can  divide  E.  Therefore  E  is  the  greatest  common 
divisor  of  A  and  B. 

The  greatest  common  divisor  of  three  expressions,  A,  B, 
C,  may  be  obtained  as  follows : 

First  find  D,  the  G.  C.  D.  of  any  two  of  them,  say  of  A 
and  B ;  next  find  F,  the  G.  C.  D.  of  D  and  C;  then  F  will 
be  the  G.  C.  D.  of  A,  B,  C.  For  D  contains  every  factor 
which  is  common  to  A  and  B  (Art.  72) ;  and  as  F  is  the 
G.  C.  D.  of  D  and  C,  it  contains  every  factor  common  to  D 
and  C,  and  therefore  every  factor  common  to  A,  B,  and  C. 
Hence  F  is  the  G.  C.  D.  of  A,  B,  C. 

EXAMPLES. 

1.   Find  the  G.  C.  D.  of  x2-ix  -f-  3  and  4a;3-  9x2-  15a;-f-18. 
x2  -  4a  +  3)4a;3  -    9a?  -  15a  +  18 (4a  +  7 
4a3  -  16a2  +  12a 


7a2  -  27a  +  18 

7a2  -  28a  +  21 

a  -    3)  a2  —  4a  +  3(a  - 

-  1 

a2  -  3a 

-    a  +  3 

—    a  +  3 

Therefore  the  G.  C.  D.  is  a  -  3. 

Explanation.  —  First  arrange  the  given  expressions  according  to 
descending  powers  of  x.  Take  for  dividend  that  expression  whose 
first  term  is  of  the  higher  degree;  and  continue  each  division  until 
the  first  term  of  the  remainder  is  of  a  lower  degree  than  the  first  term 
of  the  divisor.  When  the  first  remainder,  x  —  3,  is  made  the  divisor, 
we  put  the  first  divisor  to  the  right  of  it  for  a  dividend,  and  after 
obtaining  the  new  quotient,  x  —  1,  we  have  nothing  for  a  remainder. 
Hence,  as  in  Arithmetic,  the  last  divisor,  x  —  3,  is  the  G.  C.  D. 
required. 


116 


EXAMPLES. 


2.    Find   the   G.  C.  D.  of   8a;8 
4a;3  -  3x2  -  24a;  -  9. 

x    4a;3 

4x3 


2x2 


53a;  -  39    and 


3a;2  - 

-  24a;  -  9 

8a;3  -  2a;2  -  53a;  -  39 

5a;2  - 

-  21a; 

8a;3  -  Ga-2  -  48a;  -  18 

2x-  - 

-     3a;  —  9 

4a;2  -     5a;  -  21 

2x2  - 

-     Gx 

4a;2  -    6a;  -  18 

3a  -  9 

x  —    3 

3a;  -  9 

2x 


Therefore  the  G.  C.  D.  is  x  -  3. 

Explanation.  —  First  arrange  the  given  expressions  according  to 
descending  powers  of  x.  The  expressions  so  arranged  having  their 
first  terras  of  the  same  order,  we  take  for  divisor  that  whose  highest 
power  has  the  smaller  coefficient,  and  arrange  the  work  in  parallel 
columns,  as  above.  (1)  At  the  first  division  we  put  the  quotient  2  to 
the  right  of  the  dividend.  (2)  When  the  first  remainder  4a-  —  5.r  —  21 
is  made  the  divisor  we  put  the  quotient  x  to  the  left  of  the  dividend. 
(3)  When  the  second  remainder  2x2  —  3x  —  9  is  made  the  divisor  we 
put  the  quotient  2  to  the  right  of  the  dividend.  (4)  When  the  third 
remainder  x  —  3  is  made  the  divisor  we  put  the  quotients  2x  and  3  to 
the  left  of  the  dividend,  and  so  on. 

This  method  is  used  only  to  determine  the  compound  factor 
of  the  G.  C.  D.  Simple  factors  of  the  given  expressions 
must  first  be  separated  from  them,  and  the  G.  C.  D.  of  these, 
if  they  have  any,  must  be  reserved  and  multiplied  into  the 
compound  factor  obtained  by  the  rule. 

3.  Find  the  G.  C.  D.  of  6a;4  -  2Ga;3  +  4()ar  -  42a;  and 
18a;4  +  3x3  -  132a;2  +  63a;. 

We  have 

6a;4-26x3+  46a;2 
and 

18a;4+  3a;8- 132a;2 -(-63a; 


42a;=2a;(3a;3-13a;2+23a;-21). 


(1) 

3a;(6a;3-r-     a;2-44a;+21).     .   (2) 

The  simple  factor  2  is  found  in  the  first  expression  and 

not  in  the  second  ;    therefore  it  forms  no  part  of  the  G.  C. 

D.,  and  may  be  rejected.      Likewise  the  simple  factor  8, 

occurring  in  the  second  expression  ami  not  in  the  first,  may 


EXAMPLES. 


117 


be  rejected  as  forming  no  part  of  the  G.  C.  D.  But  the 
simple  factor  x  is  common  to  both  expressions,  and  is 
therefore  a  factor  of  the  G.  C.  D.  and  must  be  reserved. 
Rejecting  therefore  the  simple  factors  2  and  3  as  forming  no 
part  of  the  G.  C.  D.,  and  reserving  the  common  factor  x  as 
forming  a  part  of  the  G.  C.  D.,  and  arranging  in  parallel 
columns,  we  have 


3a;3  -  13a;2  +  23a; 


21 


6a;3 
6a;3 

+ 

X2 

26a;2 

+ 

44a; 
46a; 

+  21 
-  42 

27a;2 

- 

90a;  +  G3 

The  first  division  ends  here,  since  27a;'2  is  of  a  lower  degree 
than  3a;3.  If  we  now  make  27a;'2  —  90a;  +  63  a  divisor  we 
find  that  it  is  not  contained  in  3a;3  —  13a;'2  -+-  23a;  —  21  with 
an  integral  quotient.  But,  noticing  that  27a;2  —  90a;  +  G3 
may  be  written  in  the  form  9  (3a;2  —  10a;  +  7),  and  remem- 
bering that  the  G.  C.  D.  we  are  seeking  is  contained  in  the 
remainder  9  (3a;2  —  10a;  +  7),  and  that,  since  the  two  expres- 
sions 3a;3  -  13a;2  +  23a;  -  21  and  6.c3  +  a;'2  -  44a;  +  21 
have  no  simple  factors,  therefore  their  G.  C.  D.  can  have 
none,  we  conclude  that  the  G.  C.  D.  must  be  contained  in 
the  factor  3a;2  —  10a;  +  7,  and  that  therefore  we  can  reject 
the  simple  factor  9,  and  go  on  with  the  divisor  3a;2  —  10a;  +  7. 
Resuming  the  work,  we  have 


3a;3  -  13a;2  +  23a-  -  21 
3a;3  -  10a;'2  +  7a; 

3a;2  —  10a;  +  7 
3a;2  —  7a; 

—  3a;'2  +  16a;  —  21 

—  3a;2  +  10a;  —  7 

—  3a;  +  7 

-  3a;  +  7 

2)  6a;  -  14 

3a;  —     7 

Therefore  the  G.  C.  D  is  a;(3a;  -  7). 

The  factor  2  was  removed  for  the  same  reason  as  the 
factor  9. 

4.  Find  the  G.  C.  D.  of  2a;3  +  a;2  -  x  -  2  (1)  and 
3a;3  -  2a;2  +  »  -  2  (2). 


118 


EXAMPLES. 


As  the  expressions  stand  neither  can  be  divided  by  the 
other  without  obtaining  a  fractional  quotient.  This  difficulty 
cannot  be  obviated  by  removing  a  simple  factor,  since  neither 
expression  contains  a  simple  factor.  We  may  however  intro- 
duce a  suitable  factor  into  either  expression,  just  as  in  Ex.  3 
we  removed  a  factor  when  we  could  no  longer  proceed  with 
the  division  without  a  fractional  quotient.  The  given  expres- 
sions (1)  and  (2)  have  no  common  simple  factor,  therefore 
their  G.  C.  D.  can  have  no  simple  factor,  and  hence  cannot 
be  affected  if  we  multiply  either  of  them  by  any  simple 
factor. 

Multiplying  (2)  by  2  and  taking  it  for  dividend,  we  have 


-2a! 


17a; 


11 


2a;3  +   a;2  - 

7 

x-    2 

6a;3-   4a;2  +  2x  -    4 
6x3+     3a;2—  3a;-  6 

14a;3  +  7a;2- 
14a;3-  10a;2- 

7x  —  14 
4a; 

—   7a;2  +  5a; -f-  2 
17 

17a;'2- 
17a;2  — 

3a;-  14 
17a; 

—  119a;2  +  85a; +  34 

-  1 19a;2  +  21a-  +  98 

14a;  -  14 
14a;  -  14 

64)  64a; -64 
x-    1 

Therefore  the  G.  C.  D.  is  x  -  1. 


In  this  example,  after  the  first  division  the  factor  7  is 
introduced  because  the  first  remainder  —7a;'2  +  5a;  +  2  will 
not  divide  the  first  divisor  2a;3  +  a;2  —  a;  —  2. 

After  the  second  division  the  factor  17  is  introduced 
because  the  second  remainder  17a;2  —  3a;  —  14  will  not 
divide  the  second  divisor  —  7ar  -f  5x  +  2.  Finally  the 
factor  64  is  removed  as  explained  in  Ex.  3. 

Note. —The  difference  between  the  Algebraic  G.  C.  D.  and  the 
Arithmetic  G.  C.  D.  can  be  seen  by  an  example. 

Factor  (1)  and  (2)  of  last  example  as  follows: 
2a;3  +    x*  -  x  -  2  =  (x  -  l)(2a-2  +  8a  +  2), 
and    3a;8  -  2a;2  +  x  —  2  =  (x  —  1)  (3a;2  +    x  +  2). 


EXAMPLES.  119 

Now  since  the  G.  C.  D.  of  these  expressions  is  x  —  1, 
the  factors  2x2  -f  3.«  +  2,  and  3ar  +  x  +  2,  have  no  common 
factor.     But  if  we  put  x  —  4,  then 

2a;3  +    x2  -  x  -  2  =  138, 
and  3xs  -  2x2  +  x  -  2  =  162, 

and  the  G.  C.  D.  of  138  and  162  is  6,  while  3  is  the 
numerical  value  of  the  Algebraic  G-.  C.  D.,  as  —  1.  Thus 
the  numerical  value  of  the  Algebraic  G.  C.  D.  does  not  agree 
with  the  numerical  value  of  the  Arithmetic  G.  C.  D. 

The  reason  may  be  explained  as  follows  ;  the  expressions 
2X2  +  3a;  +  2  and  3»2  +  x  +  2  have  no  Algebraic  common 
factor ;  but  when  x  =  4  they  become  equal  to  46  and  54 
respectively,  and  therefore  have  a  common  Arithmetic  factor 
2,  which,  multiplied  into  x  —  1  or  3,  gives  6  for  the  nu- 
merical value  of  the  Arithmetic  G.  C.  D.,  while  3  is  the 
numerical  value  of  the  Algebraic  G.  C.  D.  In  the  same  way 
it  may  be  shown  that  if  we  give  particular  numerical  values 
to  the  letters  in  any  two  expressions,  and  in  their  Algebraic 
G.  C.  D.,  the  numerical  value  of  the  G.  C.  D.  is  by  no 
means  necessarily  the  Arithmetic  G.  C.  D.  of  the  values  of 
the  expressions. 

We  may  now  enunciate  the  rule  for  finding  the  greatest 
common  divisor  of  two  compound  Algebraic  expressions. 

Rule. 

Arrange  the  given  expressions  according  to  the  descending 
poivers  of  the  same  letter.  Divide  that  expression  which  is 
of  the  higher  degree  by  the  other;  or,  if  both  are  of  the  same 
degree,  divide  that  whose  first  term  has  the  larger  coefficient 
by  the  other;  and  if  there  is  no  remainder  the  first  divisor 
will  be  the  required  greatest  common  divisor. 

If  there  is  a  remainder  divide  the  first  divisor  by  it,  and 
continue  thus  to  divide  the  last  divisor  by  the  last  remainder, 
until  a  divisor  is  obtained  which  leaves  no  remainder;  the 
last  divisor  ivill  be  the  greatest  common  divisor  required. 


120         LEAST   COMMON  MULTIPLE  —  DEFINITIONS. 

Note  1.  —  Before  beginning  the  division,  all  simple  factors  of  the 
given  expressions  must  be  removed  from  them,  and  the  greatest 
common  divisor  of  these  must  be  reserved  as  a  factor  of  the  (Jr.  C.  D. 
required.     (See  Ex.  3.) 

Note  2.  —  Either  of  the  given  expressions  or  any  of  the  remain- 
ders may  be  multiplied  or  divided  by  any  factor  which  does  not  divide 
both  of  the  given  expressions.     (See  Ex.  4.) 

Note  3.  —  Each  division  must  be  continued  until  the  remainder  is 
of  a  lower  degree  than  the  divisor. 

Find  the  G.  C.  D.  of 

5.  x3  +  2x2  —  13a;  +  10,  and  x8  +  x°-  —  10a;  4-  8. 

Ans.  x2  —  3a;  +  2. 

6.  a;3-5a-2-99a;-f-40,  and  a;3- Ga;2- 86a;  +  35.    ^2-13x'+5. 

7.  xs-x2—5x—  3,  and  a;3—  4ar—  11a:  —  6.         a;2+  2a;  +  1. 

8.  a;3  +  3a;2  -  8a;  -  24,  and  a-3  +  3a;2  -  3a;  -  9.         x  +  3. 

9.  2a;34-4a2-7.f-14,andGa;3-10a-2-21a;+35.        2ar-7. 

LEAST    COMMON    MULTIPLE. 

75.  Definitions.  —  A  Multiple  of  an  expression  is  any 
expression  that  can  be  divided  by  it  exactly. 

Hence,  a  multiple  of  an  expression  mast  contain  all  the 
factors  of  that  expression.     Thus, 

Ga2b  is  a  multiple  of  3  or  2  or  G  or  a  or  h. 

A  Common  Multiple  of  two  or  more  expressions  is  an 
expression  that  can  be  divided  by  each  of  them  exactly ; 
or,  it  is  one  of  which  all  the  given  expressions  are  factors. 

Thus,  the  expression  aire3  is  a  common  multiple  of  the 
expressions,  a,  6,  c,  ah,  abc,  ab2,  6ac8,  etc.,  or  of  the  expres- 
sion itself ;  but  it  is  not  a  multiple  of  a2,  nor  of  b3,  nor  of 
any  symbol  which  does  not  enter  into  it  as  a  factor. 

The  Least  Common  Multiple  *  of  two  or  more  Algebraic 
expressions  is  the  expression  of  least  degree  which  is  divisi- 
ble by  each  of  them  exactly. 

*  Called  alHo  lowest  common  multiple.  The  term,  Uaat  common  multiple,  i* 
objected  to  l>y  some,  for  a  reason  similar  to  tho  ouc  for  which  they  object  to  tho 
term  greatest  common  dioisor. 


LEAST   COMMON  MULTIPLE   OF  MONOMIALS.         121 

Hence,  the  least  common  multiple  of  two  or  more  expres- 
sions is  the  product  of  all  the  factors  of  the  expressions,  each 
factor  being  taken  the  greatest  number  of  times  it  occurs  in 
any  of  the  expressions. 

The  abbreviation  L.  C.  M.  is  often  used  instead  of  the 
words  least  common  multiple. 

Note.  —  Two  or  more  expressions  can  have  only  one  least  common 
multiple,  while  they  have  an  indefinite  number  of  common  multiples. 

76.  The  Least  Common  Multiple  of  Monomials, 
and  of  Polynomials  which  can  be  easily  Factored. 

—  Let  it  be  required  to  find  the  least  common  multiple  of 
2la*x3y,  35aVy,  28a3x2y*,  and  lia5x2y2. 

By  separating  each  expression  into  its  prime  factors,  we 
have  2\aixhj    =  3    X  7a4arty, 

S5a2x*y    =  5    x  IdWy, 
28a3x2yi  =  22  x  7a3x2y4, 
14a&x2y2  =  2    x  7a5arfy2- 
Hence,  the  L.  C.  M.  =  7    x  3  X  5  x  22a5afyi 
=  42Ua5.i'Y; 
for  420  is  the  numerical  L.  C.  M.  of  the  coefficients  ;  a5  is 
the  lowest  power  of  a  that  is  divisible  by  each  of  the  quan- 
tities a4,  a2,  a3,  a5;  x*  is   the   lowest   power  of   x   that   is 
divisible  by  each  of  the  quantities  a,-8,  a;4,  x2 ;  and  yi  is  the 
lowest  power  of  y  that  is  divisible  by  each  of  the  quantities 
y,  y*,  y2. 

2.  Find  the  L.  C.  M.  of  Gxs(a  -  x)2,  8a2(a  -  x)3,  and 
\2d2x2(a  —  a;)4. 

Resolviug  into  factors,  we  have 

6x3(a  -  x)2  =  3  X  2x3(a  -  x)\ 
8a2(a  -  x)3  =  2  X  2  X  2a2(a  -  x)3, 
12a2x2(a  -  xY  =  3  x  2  x  2a2x2(a  -  x)\ 
Hence,  the  L.  C.  M.  =  3  X  23a2x3(a  -  a;)4 
=  2ia2x3(a  -  a;)4. 
For  it  consists  of  the  product  of  (1)  the  numerical  L.  C.  M. 
of  the  coefficients,  and  (2)  the  lowest  power  of  each  factor 


122 


EXAMPLES. 


which  is  divisible  by  every  power  of  that  factor  occurring  in 
the  given  expressions. 

3.    Find   the    L.    C.    M.    of    3a2   +    9a6,    2a?   -    18ab2, 
a3  +  6a26  +  dab2,  a3  -f  ba2b  +  6a62. 
Resolving  into  factors,  we  have 

3a'2  +     9a6    =  3a  (a  +  3b), 
2a3  -  18a62  =  2a(a  +  36)  (a  -  36), 
a3  +  6a26  +     9a62  =     a  (a  +  36) 2, 
a3  +  oa26  +     Gab2  =    a  (a  +  36)  (a  -f  26). 
Hence  the  L.  C.  M.  =  Ga(a  +  36)2(a  -  36)  (a  +  26). 
Hence  the  following 

Rule. 

Resolve  each  expression  into  its  prime  factors,  and  take 
the  product  of  cdl  the  factors,  giving  to  each  factor  the  highest 
exponent  which  it  has  in  the  given  expressions. 

If  the  expressions  are  prime  to  each  other,  their  product 
is  the  least  common  multiple. 


EXAMPLES. 

Find  the  least  common  multiple  of 

1.  5aa6c8,  4a62c. 

2.  12a6,  8xy. 

3.  2a6,  36c,  4ca. 

4.  a26c,  62ca,  c2ab. 

5.  5a2c,  Gc62,  36c2. 

6.  x2,  x2  -  3a. 

7.  21a3,  7a-2 (a  +1). 

8.  a2  +  ab,  ab  +  b2. 

9.  Ga2  —  2x,  9a2  —  3a. 
x2  +  2x,  x2  +  3a  +  2. 
x2  +   la  +  4,x2  + 


10. 
11. 
12. 


+  6. 


Ans.  20aW 

24abxy 

12abc 

a262c2 

30aW 


a2  +  x  —  20,  x2 


x2(x  -  3) 
21a3 (a  +  1) 
ab(a  +  b) 
6a*3a  -  1) 
a(a  +  2)(»  +  1) 
(,•  +  2)2(a  +  3) 
10a  +  24,  a'J  -  x  -  30. 
Ans.  (x  -f-  5)  (a  —  4)(«  —  6) 


LEAST   COMMON   MULTIPLE   OF  POLYNOMIALS.       123 


X 

2a;4 +  x3-  20a;2-  Ix +  24 
2a;4  +  7a;3            —9x 

2a;4+3a;3-13a;2-7a;+15 

2a;4 +  x3-20x2-lx+2± 

3 

-6x3-20x2+2x+24: 
-6x3-2lx2         +27 

2a;3 +   lx2         -   9 
2a;3  +  4a;2-6a; 

x2+2x-  3 

3a;2+6a;—  9 
3a;2+6a;-   9 

77.  "When  the  Given  Expressions  cannot  be 
Resolved  into  Factors  by  Inspection.  —  To  find  the 
least  common  multiple  of  two  compound  Algebraic  expres- 
sions in  such  cases,  the  expressions  must  be  resolved  by 
findiug  their  G.  C.  D. 

1.  Find  the  L.  C.  M.  of  2x*  +  x3  -  20a;2  -  7a;  +  24  and 
2a;4  +  3a;3  -  13a;2  -  lx  +  15. 

We  first  find  their  G.  C.  D. 

1 


■2x 


.-.     G.  C.  D.  =  x2  +  2a;  -  3. 

Hence,  by  division,  we  obtain 

2x*+  x3-  20a;2-  7a;  +  24  =  (a;2+  2a;  -  3)  (2x2-  Sx  -  8) , 
and 

2xi+Bx3-  13a;2-  7a;  +  15  =  (a;2+  2a;  -  3)  (2a;2-    x  -  5). 

Therefore 

the  L.  C.  M.  =  (a;2  +  2a;  -  3) (2a;2  -  Sx  -  8)(2a;2  -  x  -  5). 

"We  may  now  prove  the  rule  for  finding  the  least  common 
multiple  of  any  two  compound  Algebraic  expressions. 

Let  A  and  B  denote  the  two  expressions,  F  their  greatest 
common  divisor,  and  M  their  least  common  multiple.  Sup- 
pose that  a  and  b  are  the  respective  quotients  when  A  and 
B  are  divided  by  F ;  then 

A  =  aF,     and     B  =  bF.     .     . 


...   (1) 

Since  F  contains  all  the  factors  common  to  A  and  B,  the 
quotients  a  and  b  have  no  common  factor,  and  therefore 
their  least   common   multiple   is   ab,   and   hence   the   least 


124      LEAST   COMMON  MULTIPLE   OF  POLYNOMIALS. 

common  multiple  of  aF  and  bF,  or  of  A  and  JS,  from  (1)  is 
abF,  by  inspection.     That  is 

M  =  abF (2) 

But  from  (1)  we  have 

AB  =  abFF 

=  MF,  from  (2) (3) 

M  =  4*. 
F 

Rule. 

The  least  common  multiple  of  two  expressions  may  be  found 
by  dividing  their  product  by  their  G.  C.  D.,  or,  by  dividing 
either  of  the  expressions  by  their  G.  C.  D.  and  multiplying  the 
quotient  by  the  other. 

From  (3)  we  see  that  the  product  of  any  two  expressions 
is  equal  to  the  product  of  their  G.  C.  D.  and  L.  C.  M. 

To  find  the  least  common  multiple  of  three  expressions,  A, 
B,  C.  First  find  M,  the  L.  C.  M.  of  A  and  B.  Next  find 
N,  the  L.  C.  M.  of  M  and  C;  then  N  will  be  the  required 
L.  C.  M.  of  A,  B,  G. 

For  N  is  the  expression  of  least  degree  which  is  divisible 
by  M  and  C,  and  M  is  the  expression  of  least  degree  which 
is  divisible  by  A  and  B.  Therefore  N  is  the  expression  of 
least  degree  which  is  divisible  by  all  three. 

In  a  similar  manner  we  may  find  the  L.  C.  M.  of  four 
expressions. 

Note.  —  The  theories  of  the  greatest  common  divisor  and  of  the 
least  common  multiple  are  not  necessary  for  the  subsequent  chapters 
of  the  present  work,  and  any  difficulties  which  the  student  may  find 
in  them  may  be  postponed  till  he  has  read  the  Theory  of  Equations. 
The  examples  however  attached  to  this  chapter  should  be  carefully 
worked,  on  account  of  the  exercise  which  they  afford  in  all  the 
fundamental  processes  of  Algebra. 


EXAMPLES. 


125 


EXAMPLES. 


Resolve  iuto  factors 

1.  x2  +  xy. 

2.  xs  —  x2y. 

3.  10a;3  —  2oxiy. 

4.  x8  —  x2y  +  xy2. 

5.  3a4  -  3a36  +  6a262. 

6.  38a3x5  +  57aV. 

7.  aa;  —  6a;  —  az  +  6«. 

8.  2a«  +  ay  +  26a;  +  6?/. 

9.  6x2  -f-  3a;?/  —  2ax  —  ay. 

10.  2x4  —  x3  +  4x  -  2. 

11.  3a:3  +  5a;2  +  3a  +  5. 

12.  a;4  +  a;3  +  2a;  +  2. 

13.  2/8  -  y2  +  y  -  1. 

14.  2aa;2  +  Baxy  —  2bxy  —  Sby' 

15.  a2  -  19a:  +  84. 

16.  x2  -  19a;  +  78. 

17.  a2  -  14a6  +  4962. 

18.  a2  4-  5a6  +  6b2. 

19.  m2  —  13mw  +  40n2. 

20.  m2  -  22mn  +  Won2. 

21.  x2  -  23xy  +  132?/2. 

22.  130  +  Blxy  +  x2y2. 

23.  132  -  23a;  +  x2. 

24.  88  +  19a;  +  a;2. 

25.  65  +  Sxy  -  xhf. 

26.  a;2  +  16a;  -  260. 

27.  a;2  -  11a;  —  26. 

28.  a2b2  -  Babe  -  10c2. 

29.  a;4  -  a2x2  -  132a4. 

30.  4a;2  +  23a;  +  15. 

31.  12a;2  --2Sxy  +  10?/2. 

32.  8a;2  -  38a;  +  35. 

33.  12a;2  -  31a;  -  15. 


Ans.  x(x  +  y). 

x2(x  -  y). 

5a;3  (2  —  Sxy). 

x(x2  —  xy  +  y2). 

Ba2(a2  -  ab  +  2b2). 

19a8a;2(2a;3  +  3a). 

(a  —  b)  (x  —  z) . 

(2x  +  y)(a  +  b). 

(2x  +  y)(Bx  -  a). 

(2x  -  l)(a;3  +  2). 

(3a:  +  5)(a;2+  1). 

(x+  l)(a;3  +  2). 

(!/-  l)0/2+l). 

.     (2a;  +  By) (ax  -  by). 

(*-  12)  (x-  1). 

(x  -  13)  (a;  -  6). 

(a  —  lb)  (a  —  lb) . 

(a  +  36)  (a  +  26). 

(m  —  8m)  (m  —  5«) . 

(m  —  15n)(m  —  In). 

(x-  12y)(x-  Uy). 

(26  +  asy)(5  +  a?y). 

(12  -  a;)  (11  -  a). 

(8  +aj)(ll  +  •»). 

(5  +  a^)(13  -  xy). 

(x  +  26)  (a;  -  10). 

(x  +  2)  (as  -  13). 

(«6  +  2c)  (ab  —  5c). 

(a;2  +  11a2)  (a;2  -  12a2). 

(as  +  5)(4a>  +  3). 

(3a;  -  2y)(4x  -  by). 

{2x-  7)  (4a;  -  5). 

(12a;  +  5)  (a;  -  3). 


126  EXAMPLES. 

34.  3  +  11a;  -  4a;2.  Ans.  (3  -  a;)(l  +  4a;). 

35.  6  +  5a;  -  6x2.  (2  +  Sx)  (3  -  2a;). 

36.  4  -  5a;  -  6a;2.  (4  +  3a;)  (1  -  2x). 

37.  5  +  32a;  -  21a;2.  (1  +  7a;)  (5  -  3a-). 

38.  20  -  9a;  -  20a;2.  (5  +  4a;)  (4  -  5a;). 

39.  25  -  64a;2.  (5  +  8a?)  (5  -  8a;). 

40.  81pV  -  25Z>2.  (9/rz3  +  ob)(0p2zs  -  56). 

41.  (1811)2  -  (689)2.  2500  x  1122  =  2805000. 

42.  (8133)2  -  (8131)2.  16264  x  2  =  32528. 

43.  (24a;  +  y)2  -  (23a;  -  y)2.  47a;(.r  -f  2y). 

44.  (5a;  +  2y)2  -  (3a;  -  y)2.  (8a;  +  y)(2x  +  By). 

45.  9a;2  -  (3a;  -  by)0-.  by (6a-  -  by). 

46.  16a;2  -  (3a;  +  l)2.  (7a;  +  \){x  -  1). 

47.  a6  +  72963.  [a2  +  9&)  (a4  -  9a26  +  81&3). 

48.  x3ys  -  512.  (a-.y  -  8)(x2y2  +  8xy  +  64). 

49.  b00x2y  -  20y3.  20y(bx  +  y)(bx  -  y). 

50.  («  +  6)4-l.  [(«+6)2+i](ft+^  +  i)(a  +  ^_  i). 

51.  (c  +  cl)3  +  (c  -  cZ)3.  2c(c2  +  3cp). 

52.  a;4?/  —  a;2?/3  —  xhf  +  a-?/4.     .r?/(.r  +  y)  (x  —  y)  (x  -  y) . 

53.  a;4  —  6x2y2  +  y*.     (a;2  +  2xy  —  y'2)  (x2  —  2xy  -  y2). 

54.  ab(x2  +  1)  +  x(a2  +  b2).  (ax  +  &)(6as  +  a). 

55.  a3  +  (a  +  &)u.e  +  bx2.  (a2  +  &.r)  (a  +  x). 

Find  the  greatest  common  divisor  of 

56.  66aW,  44a  W,  24a263c4.  2a  W. 

57.  a;2  +  x,  (x  +  l)2,  x3  +  1.  x  +  1. 

58.  a;8  +  8y8,  a;2  +  xy  —  2y2.  x  +  2y. 

59.  12a;2  +  x  -  1,  15a;2  +  8a;  +  1.  3a;  +  1. 

60.  2a;2  +  9a;  +  4,  2a;2  -f  11a?  +  5,  2x2  -  Bx  -  2.  2x  +  1 . 

61.  Bx*— 3a8— 2a?a— x— 1,  6a;4-3a;3-a-2-a;-l.  ;ir-+l. 

62.  2a^  -  9aa;2  +  9a2a;  -  7a3,  4a;3  -  20</a;2  +  20a2.r  -  16a8. 

Ans.  x2  —  ax  ■+■  or. 

63 .  4a;5  + 1 4a;4 + 20a;3  +  70a;2,  8a;7  +  28a;6  -  8a;5 -  1 2x4  +  56a* 

Ans.  2a--(2.r  +  7). 

64.  35a,-8  +  47a;2  +  lBx +  1,  42a;4  +  41a;3  -  9a--  -  9a;  -  1. 

J.us.  7a?  +  8a;  -+-  !>• 


EXAMPLES.  127 

Find  the  least  common  multiple  of 

65.  35cm,*2,  42a/,  30az2.  Ans.  210ax2y2z2. 

66.  x2  -  3a;  +  2,  x2  -  1.  (x  +  1)  (a  -  l)(x  -  2). 

67.  or  -  ox  +  4,  a;2  -  Gx  +  8.  (a;  —  4)  (a?  -  1)  (x  —  2) . 

68.  a;2  -  x  -  6,  x2  +  a;  -  2,  a;2  -  4a;  +  3. 

Ans.  (x  -  3)  (a;  +  2)  (a;  —  1). 

69.  3a;2  +  11a;  +  6,  3a;2  +  8a;  +  4,  a;2  +  5a;  +  6. 

Ans.  (x  +  2)  (a;  +  3)  (3a;  +  2). 

70.  5a;2  +  11a;  +  2,  5a;2  +  16a;  +  3,  a;2  +  5a;  +  6. 

Ans.  (x  +  2)  (a;  +  3)  (5a;  +  1). 

71.  3a;2  -  x  -  14,  3a;2  -  13x  +  14,  x2  -  4. 

Ans.  (x  +  2)(x  -  2)  (3a;  -  7). 

72.  a;2  -  1,  x3  +  1,  a;3  -  1.  a;6  -  1. 

73.  x2  —  1,  x2  +  1,  xi  +  1,  a;8  -  1.  x8  -  1. 

74.  x2  -  1,  xs  +  1,  xs  -  1,  a;6  +  1.  a;12  -  1. 

75.  a;3 + 2a;2 - 3a;,  2a;3 + 5a;2 -3a;.      a;(a;-l)  (a;+3)  (2a;— 1). 

76.  a;2  -  10a;  +  24,  a;2  -  8a;  +  12,  a;2  -  6a;  +  8. 

Ans.  (x  -  4)  (a;  -  6)  (a-  -  2). 

77.  a;3  -r  y3,  x3y  -  y\  y2{x  —  ?/)2,  a;2  +  xy  +  y'2. 

Ans.  y'2(x  —  y)2(x2  +  xy  +  ?/2). 

78.  a2  -  62,  a3  -  63,  a3  -  a2b  -  ah2  -  2b3. 

Ans.  (a  +  6)  (a  —  6)(a  —  2?>)(«2  +  aft  +  &2). 

79.  21a?(ajy  -  y2)2,  35(x*y2  -  a;2?/4),  15?/(a;2  +  xy)2. 

Ans.  105x2y2(x  +  y)2(x  —  y)2. 


128     A    FRACTION  —  ENTIRE  AND   MIXED    QUANTITIES. 


CHAPTER     VIII. 

FRACTIONS. 

Note. — In  this  chapter  the  student  will  find  that  the  definitions, 
rules,  and  demonstrations  closely  resemble  those  with  which  he  is 
already  familiar  in  Arithmetic. 

78.   A  Fraction  —  Entire  and  Mixed  Quantities. 

—  A  Fraction  is  an  expression  of  an  indicated  quotient  by 
writing  the  divisor  under  the  dividend  with  a  horizontal  line 
between  them  (Art.  11).  In  the  operation  of  division  the 
divisor  sometimes  may  be  greater  than  the  dividend,  or  may 
not  be  contained  in  it  an  exact  number  of  times  ;  in  either 
case    the    quotient   is    expressed    by  means    of   a   fraction. 

Thus,  the  expression  -  indicates   cither  that  some  unit  is 

b 
divided  into  b  equal  parts,  and  that  a  of  these  are  taken,  or 
that  a  times  the  same  unit  is  divided  into  b  equal  parts,  and 
one  of  them  taken. 

In  any  fraction  the  upper  number,  or  the  dividend,  is 
called   the  numerator,   and   the  lower  number,   or  divisor, 

a 
6' 

which  is  read  a  divided  by  b,  a  is  called  the  numerator  and 
b  the  denominator,  and  the  two  taken  together  are  called  the 
terms  of  the  fraction.  Thus  the  denominator  indicates  into 
how  many  equal  parts  the  unit  is  to  be  divided,  and  the 
numerator  indicates  how  many  of  those  parts  are  to  be 
taken. 

Every  integer  or  integral  expression  may  be  considered  as 
a  fraction  whose  denominator  is  unity  ;  thus, 

a  ,    ,        a  -\-  b 

a  =  -,     a  +  b  =  — ^— . 


TO   REDUCE  A   FRACTION    TO   ITS   LOWEST    TERMS.     129 

An  entire  quantity  or  integral  quantity  is  one  which  has  no 
fractional  part ;  as  ah  or  or  —  2ab. 

A  mixed  quantity  is  one  made  up  of  an  integer  and  a 

fraction  ;  as  b  +  -. 

a 

A  proper  fraction  is  one  whose  numerator  is  less  than  its 

denominator  ;  as . 

a  -f  x 
An  improper  fraction  is  one  whose  numerator  is  equal  to 

or  greater  than  the  denominator ;  as  -  and -. 

a  a 

The  reciprocal  of  a  fraction  is  another  fraction  having  its 
numerator  and  denominator  respectively  equal  to  the  denom- 
inator and  numerator  of  the  former. 


79.  To  Reduce  a  Fraction  to  its  Lowest  Terms. 

—  Let-  denote  any  fraction,  and —  denote  the  same  frac- 

b  mb 

tion  with  its  terms  multiplied  by  m. 

Now  -  means  that  a  unit  is  divided  into  b  equal  parts,  and 
b 
that  a  of  these  are  taken (1) 

And  —  means  that  the  same  unit  is  divided  into  mb  equal 

mb 
parts,  and  that  ma  of  these  are  taken (2) 

Hence  b  parts  in  (1)  =  mb  parts  in  (2). 

1  part    in  (1)  =  m    parts  in  (2), 

and  .'•     a  parts  in  (1)  =  am  parts  in  (2), 

that  is, 
Conversely, 

Therefore,  the  value  of  a  fraction  is  not  altered  if  the 
numerator  and  denominator  are  either  both  multiplied  or  both 
divided  by  the  same  quantity. 


a 
b 

= 

ma 

mb 

ma, 
mb 

= 

a 
b 

130  EXAMPLES. 

When  both  numerator  and  denominator  are  divided  by 
all  the  factors  common  to  them,  the  fraction  is  said  to  be 
reduced  to  its  lowest  terms. 

Hence  to  reduce  a  fraction  to  its  lowest  terms  we  have  the 
following 

Rule. 

Divide  both  numerator  and  denominator  by  their  greatest 
common  divisor. 

Dividing  both  terms  of  a  fraction  by  a  common  factor  is 
called  canceling  that  factor. 

EXAMPLES. 

Reduce  the  following  fractions  to  their  lowest  terms. 
1     Qa2bc2  =  2ac 
9ab2c         36' 
The  greatest  common  divisor  Sabc  of  both  terms  is  can- 
celed. 

2>     lx2yz   _  J_ 
28x3yz2       Axz 
The  factor  7x2yz,  which  is  the  greatest  common  divisor  of 
both  terms,  is  canceled. 

„  24asc2x2  24asc2x2  Aac2 

18asx2  -  12a2*3  ~  Ga2x2(3a  -  2a)  ~  3a  -  2x 
Here  6a2x2  is  canceled  since  it  is  the  greatest  common 
divisor  of  both  terms. 

Note.  — .In  each  of  these  examples,  the  resulting  fractions  have 
the  same  value  as  the  given  fractions,  but  they  are  expressed  in  a 
simpler  form.  The  student  should  be  careful  not  to  begin  canceling 
until  he  has  expressed  both  terms  of  the  fraction  in  the  most  con- 
venient form,  by  factoring  when  necessary.     Thus, 

4      6x2  —  8xy  =  2x(Sx  —  Ay)  _  2x 
[)xy  -  \2y2  ~  3?/ (3a  -  Ay)  ~  Sy' 
Instead   of   reducing  a  fraction   to   its   lowest   terms  by 
dividing  the  numerator  and  denominator  by  their  G.  C.  D., 


EXAMPLES.  131 

we  may  divide  by  any  common  factor,  and  repeat  the  process 
till  the  fraction  is  reduced  to  its  lowest  terms.     Thus, 

24a3&2c3  _  12a26c2  _  Gac  __  2a 

36aW  ~  18a62c2       96c       3b 
Reduce  the  following  fractions  to  their  lowest  terms. 

G      8^c*  Ans.^. 

12ab2cd  3bd 

„      So?  -  Gab  3 


9. 


10. 


2,r0  -  Aab2  2b 

Ax2  -  9y2  2x  -  3y^ 

4x2  -(-  6xy'  2x 

20 (x3  -  f)      ^  A(x-y). 
5a;2  +  bxy  +  by2 

x3  -  2a;?/2  x 

x*  _  4ajy  _|-  4y*'  x2  —  2y2' 

When   the   factors    of    the   numerator   and   denominator 

cannot   be    found    by    inspection,    their    greatest    common 

divisor  may  be  found  by  the  rule  (Art.  74),  and  the  fraction 

then  reduced  to  its  lowest  terms. 

'    „  „  .,    ,         w  3a;3  -  13a:2  +  23a;  -  21 

11.    Reduce  to  its  lowest  terms  — — — — — — — . 

15a;3  —  38a;2  —  2x  +  21 

The  G.  C.  D.  of  the  numerator  and  denominator  is  3x  —  7. 

Dividing  the   numerator  and  denominator  by  3x  —   7,  we 

obtain  the  respective  quotients  x2  —  2x  +  3  and  5ar  —  x  —  3. 

Therefore 

3a;3-13a;2+23a;-21  _  (3a;  - 7)  {x2-  2x  +  3)  =  a;2-  2x  +  3 

15a;3- 38a;2- 2a; +  21       (3*  -7)  (5a;2-  x-  3)       5a;2-  x  -3 

This  example  may  also  be  solved  without  finding  the 
G.  C.  D.  by  the  rule  (Art.  74)  as  follows : 

By  2  of  Art.  74,  the  G.  C.  D.  of  the  numerator  and 
denominator  must  divide  their  sum  18a;3  —  51a;2  -f-  21a;,  that 
is,  3cc(3a5  —  7)(2»  —  1).  If  there  be  a  common  divisor  it 
must  clearly  be  3x  —  7.     Hence,  arranging  the  numerator 


132      TO  REDUCE  A  MIXED  QUANTITY  TO  A  FRACTION. 

and  denominator  so  as  to  show  3s  —  7  as  a  factor,  we  have 
the  fraction 

=  x*(3x  -  7)  -  2s(3s  -  7)  +  3  (3s  -  7) 

5s2(3s  -  7)  -  s(3s  -  7)  -  3  (3s  -  7) 

_  (3s  -  7)  (a8  -  2s  +  3)  =  s'2  -  2x  +  3 

(3s  —  7)  (5s2  —  s  —  3)        5s2  —  s  —  3 

When  either  the  numerator  or  denominator  can  readily  be 

factored  we  may  use  the  following  method  : 

12.    Reduce  to  its  lowest  terms  — -1 — — — : — — — ■ . 

7s3  —  18s2  +  6s  +  5 

The  numerator  =  x(x2  +  3s  —  4)  =  x(x  +  4)  (s  —  1). 

The  only  one  of  these  factors  which  can  be  a  common 
divisor  is  s  —  1,  since  the  denominator  does  not  contain  s, 
and  5  the  last  term  in  the  denominator  does  not  contain  4. 
(See  Art.  GO.)  Hence,  arranging  the  denominator  so  as  to 
show  s  —  1  as  a  factor, 

the  fraction  = *(«+*;  (*-l) =     *(*+*>    . 

7s'-(s—  1)— lls(s—  1)—  b(x—  1)       7s'-—  lis— 5 

Reduce  to  lowest  terms 

as  -  a2b  -  ab2  -  2bs  a  -  2b 


13. 
14. 
15. 


+  3a26  +  3a&2  +  268  a  +  25 

-  5s2  +  7s  -  3  s  -  3 


x3  —  3s  +  2  x  +  2 

4o8  +  12«2&  -  ab2  -  15&8  2a  +  56 


6a8  +  13a26  -  4a&2  -  1563  3«  +  bb 

80.    To  Reduce  a  Mixed  Quantity  to   the  Form 

of  a  Fraction.  —  Let  it  be  required  to  reduce  a  + 

b  +  c 

to  the  form  of  a  fraction. 

The  entire  part  a  =  -  (Art.  78)  =  a^b  +  ^  (Art.  79). 

tt  i      &c         a(b  +  r)    .      be 

Hence  a  -\ =     v       - — '  H 

6  +  c  b  +  c  6  +  o 

_  ab  +  ac,  -j-  be 

b  +  c        ' 


TO  REDUCE  A  FRACTION  TO  A  MIXED  QUANTITY.      133 
Hence  we  have  the  following 

Rule. 

Multiply  the  entire  part  by  the  denominator,  and  to  the 
product  add  the  numerator  with  its  proper  sign;  under  this 
sum  place  the  denominator,  and  the  result  ivill  be  the  fraction 
required. 

EXAMPLES. 

Reduce  the  following  to  fractional  forms : 

x2 


1.  a  —  x  + 

2.  a?  +  1  + 


3.  ar  —  xy  +  yi 

4.  a  +  b 


a  +  x 

4 
x  -  3' 


x  +  y 
a-  +  ft2 


Ans 

a2 
a  -\-  x 

x2  - 

2x  +  1 

X 

-  3 

Xs 

x  +  y 

262 

81.  To  Reduce  a  Fraction  to  an  Entire  or  Mixed 

Quantity.  —  Let  it  be  required  to  reduce  — to  a 

^  ■   ■*  a  +  x 

mixed  quantity. 

Performing  the  division  indicated,  we  have 

ax  +  2x2  _  x  +       x- 
a  +  X  a  +  x 

Hence  we  have  the  following 

Rule. 

Divide  the  numerator  by  the  denominator,  as  far  as  possible, 
for  the  entire  part,  and  annex  to  the  quotient  a  fraction  having 
the  remainder  for  numerator,  and  the  divisor  for  denominator ; 
it  will  be  the  mixed  quantity  required. 


134 


TO   REDUCE   FRACTIONS   TO    THEIR  L.    C.   D. 


EXAMPLES. 

Reduce  to  whole  or  mixed  quantities  the  following : 


1.    — 


24a 

7 
a*  +  3a6 


a  +  b 
36ac  +  4c 

9 
8a2  +  3& 

4a 
x2  +  3x  + 

2 

SB  ■+  3 

2x2  -  Gx  - 

-  1 

&•  —  3 

x4  +  1 

J.WS. 

3a  +  y. 

a 

2a6 
a  +  6' 

4ac  +  f. 

2a  +  » 
4a 

,         2 

'    a?  +  3" 

2x 

1 

x  —  3 

4-  1 

+  -2 

a?  -  1  a;  -  1 

82.  To  Reduce  Fractions  to  their  Least  Common 

Denominator.  —  Let  it  be  required  to  reduce  — ,  — ,  —  to 

yz   zx   xy 

equivalent  fractions  having  the  least  common  denominator. 

The  least  common  multiple  of  the  denominators  is  xyz. 
Dividing  this  L.  C.  M.  by  the  denominators,  yz,  zx,  and  xy, 
we  have  the  quotients  x,  y,  and  z,  respectively. 

By  Art.  79,  both  terms  of  a  fraction  may  be  multiplied  by 
the  same  number  without  altering  its  value;  therefore  we 
may  multiply  both  terms  of  the  first  fraction  by  x,  both  terms 
of  the  second  fraction  by  ?/,  and  both  terms  of  the  third 
fraction  by  z,  and  the  resulting  fractious  will  be  equivalent 
to  the  given  ones. 

a_  _   cm       b_  _   by_       c   _    cz 
yz       xyz      zx       xyz      xy       xyz 


Hence  — 


That  is,  the  resulting  fractions 


(IX 


xyz' 


and   —  have  the 

xyz    xyz  xyz 

same  values  respectively  as  the  given  fractious  — ,  — ,  and  — , 

yz   zx  xy 

and  they  have  the  least  common  denominator  xyz.     Hence, 


RULE   OF  SIGNS   IN  FRACTIONS.  135 

for  reducing   fractions  to  their  least  common  denominator, 
we  have  the  following 

Rule. 
Find  the  least  common  multiple  of  the  given  denominators, 
and  take  it  for  the  common  denominator;  divide  it  by  the 
denominator  of  the  first  fraction,  and  multiply  the  numerator 
of  this  fraction  by  the  quotient  so  obtained;  and  do  the  same 
toith  all  the  other  given  fractions. 

Note  1.  — It  is  not  absolutely  necessary  to  take  the  least  common 
denominator.  Any  common  denominator  may  be  used.  But  in 
practice  it  will  be  found  advisable  to  use  the  least  common  denomina- 
tor, as  the  work  will  thereby  be  shortened. 

Note  2.  —  It  frequently  happens  that  the  denominators  of  the 
fractions  to  be  reduced  do  not  contain  a  common  factor.    Thus,  tbe 

denominators  of  the  fractions  -,   -,  and  -  have  no   common    factor: 
b    d'  f 

therefore  the  least  common  denominator  of  tbese  fractions  is  bdf,  the 

product  of  all  their  denominators. 

EXAMPLES. 

Reduce  the  following  fractions  to  their  L.  C.  D. 

1  JL     _±     _JL_  Ans   —     -^-     — 

"    Ax     6b2'    12a;3'  12a;3'    12a;3'    12a;8' 

2  a  x  a2  a(x  +  a)    x(x  +  a)         a2 

x  —  a    x  —  a    as2  —  a2'         a;2  —  a2  '     a;2  —  a2  '  a;2  —  a2 

83.  Rule  of  Signs  in  Fractions.  —  The  signs  of  the 
several  terms  of  the  numerator  and  denominator  of  a  fraction 
relate  only  to  those  terms  to  which  they  are  prefixed,  while 
the  sign  prefixed  to  the  dividing  line  relates  to  the  fraction 
as  a  whole,  and  is  the  sign  of  the  fraction.     Thus,  in  the 

fraction  —  ^-^ — ,  the  sign  of  a,  the  first  term  of  the  numer- 
a  +  b 

ator,  is  +  understood,  the  sign  of  the  second  term  b  is  — , 

and  the  sign  of  each  term  a  and  b  of  the  denominator  is  +, 

while  the  sign  of  the  fraction  itself  is  — . 


136  RULE   OF  SIGNS   IN   TRACTIONS. 


The  symbol  means  the  quotient  resulting    from    the 

—  b 

division  of  —a  by  —  b;  and  this  is  obtained  by  dividing  a 

by  b,  and  prefixing  + ,  by  the  rule  of  signs  in  division  (Art. 

46). 

Therefore  —  =  +-  =  - (1) 

Also,  ^—  is  the  quotient  of  —a  divided  by  b  ;  and  this  is 

obtained  by  dividing  a  by  b,  and  prefixing  — ,  by  the  rule  of 
signs. 

Therefore  —  =  -- (2) 

b  b  K  ' 

In  like  manner,  is  the  quotient  of  a  divided  by  —  b; 

and  this  is  obtained  by  dividing  a  by  b,  and  prefixing  — ,  by 
the  rule  of  signs. 

Therefore  —  =  -^ (3) 

Hence,  we  have  the  following  rule  of  signs  : 

(1)  If  the  signs  of  both  numerator  and  denominator  be 
changed,  the  sign  of  the  whole  fraction  remains  unchanged. 

(2)  If  the  sign  of  the  numerator  alone  be  changed,  the  sign 
of  the  whole  fraction  will  be  changed. 

(3)  If  the  sign  of  the  denominator  alone  be  changed,  the 
sign  of  the  whole  fraction  will  be  changed. 

.   Or  they  may  be  stated  as  follows  : 

(1)  We  may  change  the  sign  of  every  term  in  the  numera- 
tor and  denominator  of  a  fraction  without  altering  the  value 
of  the  fraction. 

(2)  We  may  change  the  sign  of  a  fraction  by  changing  the 
sign  of  every  term  either  in  the  numerator  or  denominator. 


EXAMPLES.  137 


EXAMPLES. 

a  —  b         —  a-\-b         b  —  a 


m  —  n       —m  +  n       n  —  m 
b  —  a  —b  +  a  a  —  b 


—m  +  n 

-b  +  a 

2x 

3x 

2x  2x  2x 

Sx  3x  Sx 


x  —  x2  —x  +  x2  x2  —  x 

The  intermediate  steps  may  usually  be  omitted. 
From  Art.  36  we  have 
abmx  _  (~a)(-b)(-m)x  _  a(-b)(-m)x  _  etc_ 
pqr  (-p)qr  (~I>)(-ff)r 

That  is,  if  the  terms  of  a  fraction  are  composed  of  any 
number  of  factors,  any  even  number  of  factors  may  have 
their  signs  changed  without  altering  the  value  of  the  fraction; 
but  if  any  odd  number  of  factors  have  their  signs  changed, 
the  sign  of  the  fraction  is  changed. 

Thus     (a  -b)(b~  °>_^  -  q)(6  -  c)  =  (6  -  a)(c  -  b) 
'    (x-y)(y-z)      (y-x)(y-z)      (y-x)(z-y) 

When  the  numerator  is  a  product,  any  one  or  more  of  its 
factors  can  be  removed  from  the  numerator  and  made  the 
multiplier. 

Thus,  -^  =  ab-^-r  =  abed-    * 


a  +  b  a  -f  b  a  +  b 

Change   the   signs   of  the   following   fractions   so   as   to 
express  them  altogether  in  four  different  ways. 

4. 


a  —  b 
x  —  y 

.       b  —  a       b  —  a       a  —  b 

Ans. ,                 , 

y  —  x       x  —  y       y  —  x 

X 

—  X                 —X                   X 

y  —  z 

z  -  y     y  -  z     *  -  y 

b  —  a 

a  —  b              a  —  b              b  —  a 

a  —  b  +  c 

b 

—  c  —  cl        a  —  b  +  c       b  —  c  —  a 

abed 

a(- 

-b)(-c)d,    a(-b)cd          abed 

xyz 

xyz          '   xy(-z)  '  x(-y)(-z) 

138       ADDITION  AND   SUBTRACTION   OF  FRACTIONS. 


84.  Addition  and  Subtraction  of  Fractions.  —  Let 

it  be  required  to  add  together  -  and  -. 
c  c 

Here  the  unit  is  divided  into  c  equal  parts,  and  we  first 

take  a  of  these  parts,  and  then  b  of  them;    i.e.,  we  take 

a  +  b  of  the  c  parts  of  the  unit ;  and  this  is  expressed  by 

the  fraction  — — — . 
c 

a  j_b  _  a  +  b 

c       e  c 


Similarly 


a  _  b  _  a  —  b 
c       c  c 


Let  it  be  required  to  add  together  -  and  -. 

b  d 

w    ,  a       ad         ,    c        be 

We  have  -  =  — ,  and       =  — . 

b        bd  d       bd 

Here  in  each  case  we  divide  the  unit  into  bd  equal  parts, 
and  we  first  take  ad  of  these  parts,  and  then  be  of  them  ; 
i.e.,  we  take  ad  +  be  of  the  bd  parts  of  the  unit;  and  this 

is  expressed  by  the  fraction  — -. 

bd 

a   ,    c  _  ad  +  be 


Similarly 


b       d  bd 

a        c       ad  —  be 


b        d  bd 

Here  the  fractions  have  been  reduced  to  a  common  denom- 
inator bd.  But  if  b  and  d  have  a  common  factor,  the  product 
bd  is  not  the  least  common  denominator,  and  the  fraction 

will  not  be  in  its  lowest  terms.     To  avoid  working 

bd 

with  fractions  which  are  not  in  their  lowest  terms,  it  will  be 

found  advisable  to  take  the  least  common  denominator,  which 

is  the  least  common  multiple  of   the  denominators  of   the 

given  fractions  (Art.  82,  Note  1). 


EXAMPLES.  139 


Hence  we  have  the  following 
Rule. 


To  add  or  subtract  fractions,  reduce  them  to  the  least 
common  denominator ;  add  or  subtract  the  numerators,  and 
icrite  the  result  over  the  least  common  denominator. 


EXAMPLES. 

6  6  b 

Here  the  fractions  have  already  a  common  denominator,* 
and  therefore  need  no  reducing.     Hence  we  have 
a  +  c      a  —  c      a  +  d  _  a  +  c  +  a  —  c  +  a  +  d  _  3a  +  d 

6  6  6  b  b 

2.    Add — —  and  — . 

3a  9a 

Here  the  least  common  denominator  is  9a.      Hence  we 
have  (Art.  82) 


2x  + 
3  a 

^(  + 

bx  - 

9a 

4a 

_  3  (2x 

9 

+  «) 
a 

,  5x 

9a 

4a, 

Cu-  + 

3a  + 

").)•  — 

1, 

l 

11.x-  —  a 

9a 

9a 

3.  From 

4a  - 

c 

26 

,  .  3a 
take  — 

-  36 

c 

4a  - 

2b 

3a 

—  3 

b       4a 

-  26  - 

-  (3a 

c 

36) 

c 

c 

4a 

-  26  - 

-  3a 

+ 

36 

a  +  6 

Note  1.  —  To  insure  accuracy,  the  beginner  is  recommended  to  put 
down  the  work  in  full;  and  when  a  fraction  whose  numerator  is  not  a 
monomial  is  preceded  by  a  —  sign,  he  is  recommended  to  enclose  its 
numerator  in  a  parenthesis  as  above  before  combining  it  with  the 
other  numerators. 


140 

4.    Add 
Here 


EXAMPLES. 

a 

a 

6          b  —  a 

b 
b  -  a 

a 

b 

-(Art. 

83). 

a 

b        b  —  a 

a 

b 

a 

a  — 

b 

a  — 

b 

a  — 

a  — 

b 

b 

=  1. 

X 

— 

2y   3y  —  a 

and  — 

Sx 

xy 

ay 

ax 

5.   Add 


The  least  common  denominator  is  axy. 

x  —  2y       3y  —  a       2a  —  3.x- 
xy  ay        "~        ax 

_  a(x  -  2ij)  +  a(3?/  -  a)  +  ?/(2a  -  3a>) 
a.*:y 

a#  —  2ay  +  Sxy  —  ooj  +  2a?/  —  3xy 
axy 

since  the  terms  in  the  numerator  destroy  each  other. 

n     T.         a  +  b  .  ,      a  —  b 

6.    Irom  — ! —  take  . 

a  —  b  a  -\-  b 

The  least  common  denominator  is  a2  —  Zr. 

a  +  b  _  a  —  b  _   (a  +  ?>)'2  _  (a  —  6)2 


0, 


a  —  b        a  +  b          a'2  —  If          a-  —  lf 

a2  +  2ab  +  b2  —  (a2  - 

-  2aft  +  &2) 

a2  -  62 

4a6 

a2  -  &2' 

_      .  , ,  2as  —  3a        ,      2x  —  a 

7.    Add  -               and . 

x  —  2a                x  —  a 

The  least  common  denominator  is  (x  —  2d)  (x  —  a). 


EXAMPLES.  141 


Hence  we  have 


2x  —  3a  _  2x  -  a  _  (2x  —  3a)  (x  —  a)  —  (2x  —  a)  (x  -  2a) 
x  —  2a        x  —  a  (x  —  2a)  {x  —  a) 

_  2a;2  —  5ax  +  3ft2  —  (2a;2  —  oax  -+-  2a2) 
(x  —  2a)  (a;  —  a) 
2X2  -  box  +  3a2  -  2a;2  +  5ax  -  2a2 
(x  —  2a)  (x  —  a) 


(x  —  2a)  (x  —  a) 

Note  2.  —  In  finding  the  value  of  an  expression  like 

—  (2x  —  a)[x  —  a), 

the  beginner  should  first  express  the  product  in  parentheses,  and  then 

after  multiplication,  remove  the  parentheses,  as  we  have  done.     After 

a  little  practice  he  will  be  able  to  take  both  steps  together. 

Note  3.  —  In  practice,  the  foregoing  general  method  may  some- 
times be  modified  with  advantage.  When  the  sum  of  several  fractions 
is  to  be  found,  it  is  often  best,  instead  of  reducing  at  once  all  the 
fractions  to  their  L.  C.  D.,  to  take  two  or  more  of  them  together,  and 
combine  the  results. 

o      c       ,.-.     ft  -(-  3        «  +  4  8 

8.    Simplify  — — — . 

ft  —  4     .  ft  —  3        ft'2  —  16 

Here,  instead  of  reducing  all  the  fractions  to  the  least 

common  denominator  at  once,  we  may  take  the   first  two 

fractions  together,  as  follows  : 

q  +  3  _  0+4  _       8       _  ft2- 9  —  (ft2  —  16)  _ 

a  — 4       a- 


a2-  16         (ft  -  4)  (ft  -  3)         «2-  16 

7                             8 

(ft -4)  (« -3)       (« +  4)  (ft - 

-4) 

52  —  ft 

(a  -  4)(«  +  4)(ft-  3) 

1                 1                 2x                Ax* 

9.    Simplify 

ft  —  X         ft  +  X         ft"  +  x-         ft4  +  X* 

Here  we  see  that  the  first  two  denominators  give  L.  C.  D. 
ft2  —  a;2 ;  and  this  with  «2  -f-  x-  gives  L.  C.  D.  «4  —  x4 ;  and 
this  with  ft4  +  a;4  gives  L.  C.  D.  as  —  xs.     Hence,  instead 


142  EXAMPLES. 

of  reducing  all  the  fractious  to  the  least  common  denom- 
inator, we  proceed  as  follows  : 


The  first  two  terms 
The  first  three  terms 


The  whole  expression  = 


a  +  x  —  (a  —  x)  _        2x 


a2 

-  X'2 

as 

-  X2 

2., 

2x 

d' 

■  — 

X2 

a2 

-+-  X2 

•2. 

:((/' 

+ 

*2)- 

-  2.s(a2 

— 

X2) 

a4 

-  a4 

4x3 


a4  -  xi 

4x3 

a4  -  a;4 

8a;7 

4.x-3 


t4  +  x* 


10.   Simplify 


(a-b)(a-c)  (b-c)(b-a)  (c-a)(c-6) 
The  beginner  is  very  liable,  in  this  example,  to  take  the 
product  of  the  denominators  for  the  least  common  denom- 
inator, and  thus  to  render  the  operations  very  laborious. 
The  denominator  of  the  second  fraction  contains  the  factor 
b  —  a,  aud  this  factor  differs  from. the  factor  a  —  b,  which 
occurs  in  the  denominator  of  the  first  fraction,  only  in  the 
sign  of  each  term.  Also  the  denominator  of  the  third 
fraction  contains  the  factors  c  —  a  and  c  —  b,  which  differ 
from  the  factors  a  —  c  and  b  —  c  iu  the  denominators  of  the 
first  two  fractions,  only  in  the  signs  of  each  term.  It  is 
better  to  arrange  these  factors  so  that  a  precedes  b  or  c,  and 
that  b  precedes  c.     By  Art.  83  we  have 

b  b 


and 


(b  -  c)  (b  -  a)  (b  -  c)  (a  -  b) 

c  c 


(c  —  a)  (c  —  b)        (a  —  c)  (b  —  c) 
Hence  the  given  expression  may  be  put  in  the  form 
a  be 


(a  -  &)(a  -  c)        (6  -  c)(a  -  &)        (a  -  c)(&  -  c) 
whose  L.  C.  D.  we  see  at  once  is  (u  —  b)  (<i  —  c)  {b  —  c). 


EXAMPLES.  143 

Reducing  the  fractions  to  the  least  common  denominator, 
the  given  expression  becomes 
a(b  —  a)  —  b(a  —  c)  +  c(a  —  b) 
(a  -  6)  (a  -  c)  (b  -  c) 

_  ab  —  ac  —  ab  +  be  +  ac  —  be  _ 
(a  -  6)  (a  -  c)  (6  -  c) 
The  work  is  often  made  easier  by  completing  the  divisions 
represented  by  the  fractions. 

11.    Simplify  1  +  2x+  1  -  4*  +  5. 
2x  -  2        2a;  +  2 

We  have  by  division 

1  +  1  +  —    -  2  3  1 


2a;  -  2  2as  +  2       2jc  -  2       2a;  +  2 

_  3a;  +  3  -  x  +  1 
2ar  -  2 
a-  +  2 


a;2  -  1" 

jo     5a;  —  1        3a;  —  2       a;  —  5 

8                   7                 4 

4n3   25.,  -  61 

56 

13     2a;  +  5       x  +  3        27 

a;                2a;           8a;2 

12a;2  +  28a;  -  27 
8a? 

a;  _  4       a-  -  7 

6 

a;  —  2       a;  —  5  (a;  —  2)  (a;  —  5) 

,r     4a2  +  &2       2a  —  b  Aab 


16. 


4a2  -  62       2a  +  6  4a2  -  b' 

5  3.r  4  -  13a;  1  —  6a:5 


1  +  2a;        1-2.,        1  -  4a;2  1  -  4a;2 


7.        3       +        2        +       5a;  2(13a;  +  7) 


18. 


x  -  2       3a;  +  6       x2  -  4  3(aj  -  2)  (a;  +  2) 

3a;  2  2  7a; 


1  —  x"       x  —  1       x  -f-  1  1  —  ar 

19. h +  1 .        o 

(a-b)(a-c)       (b-c)(b-a)        (c-a)(c-b) 


144        TO  MULTIPLY  A   FRACTION  BY  AN   INTEGER. 

85.  To  Multiply  a  Fraction  by  an  Integer. — 

Rule. 

Multiply  the   numerator  by  that    integer;    or  divide  the 
denominator  by  that  integer. 

The  rule  may  be  proved  as  follows : 

(1)    Let  -  denote  any  fraction,  and  c  any  integer;    then 

will  -   X  c  —  — ;   for  in  each  of  the  fractions  -  and  —  the 
b  b  b  b 

unit  is  divided  into  b  equal  parts,  and  the  number  of  parts 


(2)    Let  —  denote  any  fraction,  and  c  any  integer  ;  then 

=  -(Art.  79). 
b 

86.  To  Divide  a  Fraction  by  an  Integer. — 

Rule. 
Divide  the   numerator  by   that   integer;    or   multiply   the 
denominator  by  that  integer. 

The  rule  may  be  proved  as  follows  : 

(1)  Let  —  denote  any  fraction,  and  c  any  integer;   then 

will i-ca-;  for  —  is  c  times  -  (Art.  85) ;   and  there- 

b  b  b  b 

fore  -  is  the  quotient  of  —  divided  by  c. 

(2)  Let  -  denote  any  fraction,  and  c  any  integer  ;  then 

-  =  —  (Art.  79). 
b        be 

a  ae  a    ,      , .  x 

•'•     t  -*-  c  =  ; — '-  °  =  7-1  by  (1). 
6  be  be     J  v  J 


TO   MULTIPLY  FRACTIONS  —  EXAMPLES.  145 

87.    To  Multiply  Fractions.  —  Let  it  be  required  to 


multiply  ^  by 

c 

Put   -  =  flJ, 

6 

and 

l-  =  y- 

Then  (Art.  85; 

a  =  6a;, 

and    c  =  dy, 

therefore 

ac  =  bdxy  ; 

divide  by  6d, 

ac 

—  =  xy. 

bd        J 

But 

a    ,  c 
b       d 

""'     6 

x   c  -  ac ' 
d       bd' 

and  ac  is  the  product  of  the  numerators,  and  bd  the  product 
of  the  denominators.     Hence  the  following 

Rule. 

Multiply  the  numerators  together  for  the  numerator  of 
the  product,  and  the  denominators  for  the  denominator  of  the 
product. 

Similarly,  the  rule  may  be  demonstrated  when  more  than 
two  fractions  are  multiplied  together. 

Note.  —  If  either  of  the  factors  is  a  mixed  quantity,  it  is  usually 
best  to  reduce  it  to  a  fractional  form  before  applying  the  rule.  Also, 
it  is  advisable  to  indicate  the  multiplication  of  the  numerators  and 
denominators,  and  to  examine  if  they  have  common  factors;  and,  if 
so,  to  cancel  them  before  performing  the  multiplication. 


EXAMPLES. 

1.    Multiply  together  -?  and  -^. 
1  J      &  46  9a 

3a        8c  =  3a  X   8c  =  2c  (AA  79)> 
46       9a       46  X  9a       36 


146  EXAMPLES. 


2.  Multiply      2a2      by  <a  +  b>>\ 

2a2  (a  +  &)2  _   2a2 (a  +  6)  (a  -f  b) 

a?-b2  4a2b  (a  +  6)  (a  -  6)4a26 

a  +  6 
26(a  -  6)' 
by  canceling  those  factors  which  are  common  to  both  numer- 
ator and  denominator. 

q     tit  u-  i    2a2  4-  3a       -.    4a2  —  6a  .       ,, 

3.  Multiply 2- —  and  together. 

J        4a3  12a  +  18     fe 

2a2  +  3a        4a2  -  6a  _  a (2a  +  3)2a(2a  -  3) 
4a8  12a  +  18  4a3(2a  +  3)6 

_  2a  -  3 
12a 

4.  Multiply  f  +  ^+  lby|  +  ^-  1. 

b       a  b       a 

«  +  I  +  1  =  ft2  +  *'  +  «&  (Art.  84), 
6       a  a& 

and  «  +  *  -  1  =  q'  +  h2  ~  ab. 

b       a  ab 

a&  a&  a2&2  J 

a4  +  fr4  +  a262 
a8&8 

Otherwise  thus : 

,2 


G+s+'XH-O-G+y 


=  «^    ,    6_2    ,    j  =  a4  +  64  +  agft8 
&2  +  a2  +  a2b2 

Simplify 
5.    ^X^X   .*-*,     4»  * 


aj  _  l       x1  -  1       (a;  +  2)2  (as  -  1)  (a  +  2) 

6.    X     j.  x  —  a. 

x  +  a       \a       xj 


TO  DIVIDE  FRACTIONS  —  EXAMPLES.  147 

Ans.-fU' 


\        a  -  b)\        a  +  b) 


b1 

x) ax 

a2  +  2ax  +  x'1  "  a2  —  2ax 


v(a  —  x)  a(a  +  a;)  ax 

—  x  — 


88.  To  Divide  Fractions. — Let  it  be  required  to  divide 

-  by  « 

b    J  d 

Denote  the  quotient  b}r  x.     Then,  since  the  quotient  mul- 
tiplied by  the  divisor  gives  the  dividend  (Art.  44),  we  have 

d       b 
Multiplying  by  -,  we  have 

xxSxd=axd 

d       c       b       c 
Therefore,  Art.  87,  and  canceling  factors  common   to  the 
numerator  and  denominator,  we  have 
_  ad 
X  ~  ~bc 

That  is,  »  +  S.  =  2*  =  <L  x  £ 

b       d        be        b        c 

Hence  the  following 

Rule. 

Invert  the  divisor,  and  proceed  as  in  multiplication. 


EXAMPLES. 

b 


1.    Divide  a  by 


a  =  -  (Art.  78). 

a_j_6__a       c  _  ac 
1    '    c  ~  1       b  ~  b  ' 


148  COMPLEX  FRACTIONS. 


2. 

Divide  - — 
(a 

-  b* 

+  by 

^d2 

b 

-  b2' 

ab  -  b2 

(a  +  by 

b 

ab  -  b2 
(a  +  by 

X 

a2  — 
•6 

lr 

a2 

-  b2 

_  Ha  - 

b)(a 

+  b)(a,-b)     . 

(a 

-by 

b(a  +  b)2  a  +  b 

Simplify 

0       Ux2  -  7x         2x  -  1 


1. 


12a8  +  24a;2 

x2  +  2. 

c 

a2b2  +  Sab 

ab  +  3 

Ad2  -  1 

2a  +  1 

a2  -  121 

a  +  11 

a2  -  4 

a  +  2 

2a2  +  13.x  -f 

•  15        2.r 

'4-H.i 

+  5 

4x2  -  9 

4a2  - 

1 

a?  _  nx  - 

15        o-2  - 

-  12a  - 

-  45 

Mm 

s.  — .. 
12 

ab 

2a 

-  1 

a  - 

-  11 

a 

-  2 

2x 

-  1 

2x 

-  3 

X 

+  1 

6. 


x*  _  4X  _  45         .t.2  _  6a,  _  27  x  +  5 

89.  Complex  Fractions.  —  A  fraction  whose  numerator 
and  denominator  are  whole  numbers  is  called  a  Simple  Frac- 
tion. A  fraction  whoso  numerator  or  denominator  is  itself  a 
fraction  is  called  a  Complex  Fraction.     Thus, 

a        a 

_,  _,  _  are  complex  fractious. 
c    b    c 

c  d 

Since  a  fraction  may  be  regarded  as  representing  the 
quotient  of  the  numerator  by  the  denominator  (Art.  78), 
a  complex'  fraction  may  be  regarded  in  the  same  way  ; 
therefore,  to  simplify  a  complex:   fraction. 

Divide  the  numerator  by  (he  denominator,  as  in  division  of 
fractions  (Art.  88). 


EXAMPLES.  140 


EXAMPLES. 


1.    Simplify  1,  -,  and  -. 
a  1  1 

b   b  b 


Here  I=l^=lX-  =  ^ 

ft  6  a       a 


a  1  7  7. 

-  =  a  -= —  =  a  x  o  =  ft0. 
1  0 


-  -  1-^1  =  Ix-  =  - 

1       a  '   b       a      1       a 


The  student  should  be  able  to  write  down  the  above  results 
readily  without  the  intermediate  steps. 


2.    Simplify 


or 

x i 

x3 


This  fraction  =  tx  +  -\  -*-  (x  -  2?) 


x2  +  a2  _^_  x* 
x 


■r2  +  ft2  x        x> 


x*  —  <r      x*  —  or 


150  EXAMPLES. 

a2  +  b2  _  a2  -  b2 
3.   Simplify    ft2  "  ^  +  b2 


a  +  b  _  a  —  b 
a  —  b       a  4-  & 
a2  +  &2  _  a2  -  U1  _  (a2  +  &2)2  -  (gg  -  fe«)a 
a2  -  &2       a2  +  b2  (a2  -  &2)  (a2  +  £2) 

_  4a262 

(a2  -  b2)(a2  +  62)' 
a  +  6  _  a  —  6  _  (a  +  ft)2  -  (a  -  &)2  4a& 

a  —  b       a  +  b  a2  —  b2  a2  —  I 

Hence  the  fraction 

4a2&2  4a& 


(a2  -  62)  (a2  +  62)        a2  -  b2 

4a2b2  _  a2  —  ft2  ab 


(a2  -  62)  (a2  -(-  b2)  4ab  a2  +  b2 

Note  1.  —  In  this  example  the  factors  a  —  b  and  a  +  b  are  multi- 
plied together,  and  the  result  a2  —  62is  used  instead  of  (a  —  b)(a  +  b). 
In  general,  however,  the  student  will  find  it  advisable  not  to  multiply 
the  factors  together  till  after  he  has  canceled  all  the  common  factors 
from  the  numerator  and  denominator. 

Note  2. — When  the  numerator  and  denominator  are  somewhat 
complicated,  to  insure  accuracy  and  neatness,  the  beginner  is  advised 
to  simplify  each  separately  as  in  the  above  example. 

Note  3.  —  The  terms  of  the  simple  fractions  which  enter  into  tha 
numerator  and  denominator  of  the  complex  fraction  are  sometimes 
called  Minor  Terms.  Thus  in  Ex.  2,  a2  and  o4  are  minor  numerators, 
and  a;  and  x3  are  minor  denominators. 

It  is  often  shorter  to  reduce  a  complex  fraction  to  a  simple 
one  by  multiplying  both  terms  of  the  fraction  by  the  least 
common  multiple  of  all  the  minor  denominators. 

x  +  5  +  2 

4.    Simplify  ® 

x       x2 
Multiplying  both  terms  by  x2  we  get 

x8  +  5x2  +  Ga;  _  a; (a;  +  2)(x  +  3)  _  a-(a;  +  3) 
x2  +  6a?  +  8    ""    (x  +  2)  (x  +  4)  a?  +  4 


A  SINGLE  FRACTION  EXPRESSED  AS   A    GROUP.      151 

3x  -  8 


5.    Simplify 


.'• 


x 


4  +  x 
In  the  case  of   Continued  Fractions,  we  begin  with  the 
lowest  complex  fraction,  and  simplify  step  by  step.     Here 
multiplying  both  terms  of  the  fraction  which  follows  x  —  1 
by  4  +  x,  the  given  fraction  becomes  at  once 
3a;  -  8  _  3a  -  8 

4  -j-  x  —  x  4 

and  now  multiplying  both  terms  by  4,  we  have 

4 (3a  -  8)  =  4(3a;  -  8)  =  ^ 

4(x  -  1)  -  (4  +  x)  3x  -  8 

Simplify 

1 
x 

6     if  Ans.  x  —  1. 

'  i  +  i 

x 

I  _  1  _  1 

„     x       ar       x3  %  +  1 


X 


90.  A  Single  Fraction  Expressed  as  a  Group  of 
Fractions.  —  Let  it  be  required  to  express  the  fraction 
bxhi  -  10ay2  +  15?/3  -  5a3 
lOxhf 
as  a  group  of  four  fractions. 

The  fraction  =  ^JL  _  _  +  _  -  — 


J_  _  1 

2y       x   '    2x2       2y 


~*~    9^.2  0„2* 


152  EXAMPLES. 

Express  each  of  the  following  fractious  as  a  group  of 
simple  fractious  in  lowest  terms. 

u    3-^  +  x,r  -  y\  Ans^  x  y  _  f 

My  3  9        9a 

2     3a3x  —  4a2a2  +  6aa3  a2  _  ax       a2 

12ax  4  I       ? 


0  be  +  ca  +  ab  1        l 

O. .  —    -I-    — 

abc  a       b 

EXAMPLES. 

Reduce  to  lowest  terms  the  following  examples  : 

1  x2  +  3a  +  2 
a-2  +  6a  +  5* 

2  x*  +  10a?  +  21 

a,2  -  2a;  -  15  ' 

3  3a,-2  +  23a  -  36 
4a2  +  33a  -  27' 

,     x2  -  lO.i-  +  21 


c 


46a 


Here  we  see  at  once  that  the  numerator  =  [x  —  1)(x 
find  by  trial  that  x  —  1  is  a  factor  of  the  denominator. 

,            x2  +  9a-  +  20  . 

5.    — L- ■ -.  Ans. 


Ans 

x  +  2 

x  -\-  5 

x  — 

7 
5 

3a-  - 

4 

4a  - 

3 

X 

—  o 

•  + 

7x  + 

3 

—  8 

);  and 

wt 

9. 
10. 


Xs  + 

7a2  +  14a  +  8 

a2  +  a  —  42 

a3  - 

10a2  +  21a  +  18' 

6a2  - 

-  11a  +  5 

3a8  - 

-  2a2  -  1" 

20a2  +  a  -  12 

12a8 

—  5a2  +  5a  -  6 

a8  - 

8oj  -  3 

a4  - 

7a-  +   1 

a8  — 

a2  -  7a  +  3 

x  +  5 

a2 

+  3a  + 
x+  7 

2 

a2 

-  4a  - 
(')./■  —  5 

3 

3a2  +  a  + 

1 

5x  +  4 

3a 

?  +  x  + 
a  -  3 

2 

a2 

-  3a-  + 

1 

x  — 

3 

a4  +  2x»  +  2x  -  1  x*  +  1 


EXAMPLES. 

11. 

X*  -   1 
X6  -    1 

12. 

tf  +  &  +  &  +  x  +  i 

a;5  -  1 

18. 

sc8  -  6a2  -  37a;  +  210 

153 

a;2  +  1 
} 

1 


ar*  +  a;2  +  1 


x  -  1 

a;  —  5 

v&  +  4ajj  _  47a;  _  210'  a;  +  5 

Reduce  to  fractional  forms  the  following  examples : 

1  a  ,         ■    «'2  —  a«  a       a2  +  x2 

14.  a  -+-  x  -\ .  Ans. 

x 

2x* 

15.  a2  —  ax  4-  x2  — 


16.    x  +  5 


a  4-  x 
2x  -  15 
x  -  3  " 


a; 

a3 

-  a;3 

a 

-fa; 

a;2 

a: 

-  3 

«5 

Reduce  to  whole  or  mixed  quantities  the  following  examples 


a  a 

19.  i^J^2.  1  _  .  -  _£_ 

1  -f  a;  1  +  a: 

20.  L£J°                                               1  4-  5a,  +  -^- 
1  -  3a;  1  -  3a; 

21.  «+7.  1  +  -$- 

aj  +  2  a;  +  2 

22.  3a;  ~  2.  3  -      17 


a;  +  5  as .  +  5 

23.  »*  -7.-1.  to  _  ,  _  ^L    ■ 

a;  —  3  x  — 

24.  *  -  2*    .  .  -  1  2*  -  X 


a;  4-  1  JB2  —  as  +  1 

25.    *L±_1.  a;3  4-  a2  +  *+!  + 


a;  —  1  a;  — 

s4  -  1 
a;  4-  1 


26. -.  x*  -  x2  4-  aj  -  1 


154  EXAMPLES. 

Perform  the  additions  and  subtractions  indicated   in  the 
following  examples : 

27.    _L_  +  .^2_S. 

x  -f  y       x-  -  y- 

3a  2ax 


+ 


a  —  x       a  -f  x       a-  —  x2 

29.  § 5 2x-  7_ 

x       2x  —  1       4a;2  —  1 

30.  _2-  +  -^_  +    *"*" 


a  —  b       a  +  b       a.    —  b4 

si.  _*_ _  .  *~3   .  +    *s 


a;  +  4       a-'2  —  4a;  +  16        x3  +  04 

„o     a-'2  +  fta;  +  a'2  _  x2  —  ax  +  a2 
xs  —  a3  x3  +  a3 

33.  32  +  ?/2  _  „_^! y2 

an/  a??/  +  y2       x2  +  an/ 

34     x*  -  2x  +  3  +       *  -  2  — 1 


^4?is.  

35-2/ 

4a 

a  -+-  a- 

2a;  -  3 

a; (4a;2  -  1) 

2a4  +  6a2&a 

a4  -  64 

2a;2  -  9x  +  44 

a;3  +  64 

2a 

35. 


a;3  +  1,  x2  -  x  +  1       x  +  1 

1  2 


a;z 

a' 
1. 

.r 

_ 

2s 

a;3 

+ 

l 

(0,-3)^-4)     (.»-2)(.c-4)     (.C-2)(z-3) 
1  ! 


4(1  +  a?)        4(1  -  a?)        2(1  +  a,-2)  1  -  x* 

37.    ^Z^ &Z^ .   (See  Art.  84^,  Ex.  10.) 

(a-b)(x-a)     (b-a)(b-x) 


(x  —  a)  (a;  —  b) 

38 1_  1 1 

(a2-b2)  (x2+  If)      (&2-a2)  (x2+a2)      (x2+a2)  (x2+  If) ' 

Ans.  0. 

39.  1  +  1 L. 

a{a  —  6)  (a  —  c)        b{b  —  a)  (b  —  c)        a6c 

1 


.4>KS'. 


c(a  —  c)  (c  —  6) 


EXAMPLES.  155 


40.    £ r  + £-_ 


41. 


(a-6)(a-c)       (6_a)(6_c)       (c-a)(c-6) 

-4ns.  1. 

1 + I + 1 . 

(a-b)(a-c)(x-a)^(b-a)(b-c)(x-b)      (C-a)(c-b)(x-c) 

1 


Ans. 


(x  —  a)(x  —  b)(x  —  c) 


Simplify  the   following   examples   in   multiplication   and 
division  : 

42>    2a-2  4-  5a-  +  2  a;2  4-  4a;      ,  ^ 


2asa  +  9a;  +  4  x  —  2 

43      2a;'2  -  x  -  1         4a;2  +  ar  —  14  *  —  1 


2a;2  +  5a;  +  2  16a;2  -  49  4a;  +  7 

44^     a;2+a;-2  ^  ^  +  5a;  +  4  .  /  a;2  +  3a-  +  2  x  x  +  3\  ^ 
a;2— a;— 20          a;2  — a;         \a;2— 2a— 15         a-2    / 

45         a;4  -  8a-  a;2  +  2a-  +  1  ^  a;2  +  2a;  +  4  j 

xi  —  4a-  —  5       a;3  —  ar  —  2x  x  —  5 

46>    (q  +  ?>)2-c2  x  a  x   (q-fe)2-^  1 

a2  +  ab  —  ac       (a-j-c)2  —  b2        ab  —  b'2  —  bc  b 
3x       x  -  1 

47 I ! 1. 


¥(*+i)  "|-  2* 


x  -  1  +       6 


x  —  6                                                       k  —  4 
4o.    .  . 

x  -  2  +  -1-  K  ~  5 

«  -  6 

49     1  ~  a*  _,_  i/     j        .    _L_\          i 

1  +  aa-)2  -  (a  +  a?)2   '  2\1  -  »       1  +  a;/' 


( 

a-2  +  i/2 


a; 


,n  #  a;2  -  w2 

50'     i_i     x  ^T7«- 


0 


as 


156  EXAMPLES. 


51.    1 7.  Ans. 


i  .  +  j 


£B 


52.    1  + 


^  1  +  x 

1+X  +  -**-'  1  +  + 

I    —   X 


53.    1 x  +  1. 

x 1_ 

i+i 

a; 


54. 


1  + 


1  +  a;2 
1  +  x 


,       2x 
1  +  x  + 


1   -  a; 


V«  +  2/      a;-?/      x'-y-J      \x  +  y      x'-y) 


*+i 

56.    -  ! 


1 y&yz  +  x  +  z) 


X 

57.    '  "  ^  v         '          x 


\2/        •'•/\.'/"  -  -''7      a2  +  an/ 


ay/  -  f        x  +  v/ 


58. 


x  —  1       x  -  1        x  +  3  _  x  +  3 

3  x  —  2  ^       7  x  +  4  28(s  +  4) 


s  +  2  +  x  +  2    '    a;  —  2  +  a:  -  2  Q(x  +  3) 

4  a;  —  3  3  x  —  1 


SOLUTION   OF  HARDER  EQUATIONS  —  EXAMPLES.    157 


CHAPTER    IX. 

HARDER     SIMPLE     EQUATIONS     OF    ONE 

UNKNOWN     QUANTITY. 

91.  Solution  of  Harder  Equations.  —  We  shall  now 
give  some  simple  equations,  involving  Algebraic  fractions, 
which  are  a  little  more  difficult  than  those  in  Chapter  VI. 
These  may  be  solved,  by  help  of  the  preceding  chapter  on 
fractions,  and  by  the  same  methods  as  the  easier  equations 
given  in  Chapter  VI. 

The  following  examples  worked  in  full  will  sufficiently 
illustrate  the  most  useful  methods. 

EXAMPLES. 

1.    Solve  6^^  =  ^-2. 
2x  +  7        x  +  5 

The  L.  C.  M.  of  the  denominators  is  (2x  +  7)(x  -+■  5). 

Clearing  the  equation   of   fractions   by  multiplying   each 

term  by  (2x  -f  7)  (x  +  5),  we  have* 

(6x  -  3)0  +  5)  =  (3a  -  2)  (2a;  +  7), 

or  6a;2  -f-  27a;  —  15  =  6.c2  +  17a;  -  14  ; 

.-.     10a;  =  1  ;         .'.     x  =  TV 

We  may  verify  this   result   by  putting   -^   for  x   in   the 

original  equation,  as  in  Chapter  VI.  ;  it  will  be  found  that 

each  member  then  becomes  — 1> 

Note  1.  — When  the  denominators  of  the  fractions  involved  con- 
tain both  simple  and  compound  factors,  it  is  frequently  best  to  multiply 
the  equation  by  the  simple  factors  first,  and  then  to  collect  tbe  integral 
terms;  after  tins  tbe  simplification  is  readily  completed  by  "multi- 
plying across  "  by  tbe  compound  factors. 

*  This  is  called  "  multiplying  across." 


158  EXAMPLES. 

2     Solve  8x  +  23  _  bx  +  2  =  2x  +  3  _  i 
20  3a;  +  4  5 

Multiplying  by  20,  the  L.  C.  M.  of  the  simple  factors  in 
the  denominators,  we  have 

Sx  -f  23  -  20(5a;  +  2)  =  8*  +  12  -  20. 
3x  +  4 

Transposing,  31  =  20{5x  +  2). 

3a;  +  4 

Multiplying  across  by  3a;  +  4,  we  have 
93a;  +  124  =  20(5a;  +  2), 
or  84  =  7x;         .-.     x  =  12. 

We  may  verify  this  result  as  before  ;  it  will  be  found  that 
each  side  becomes  -2^. 

Note  2.  —  The  student  will  see  that,  even  when  the  denominators 
of  the  fractions  contain  all  simple  factors,  it  is  sometimes  advanta- 
geous to  clear  of  fractions  partially,  and  then  to  effect  some  reduc- 
tions, before  removing  the  remaining  fractions. 

3.    Solve  X-±^  _  2x-lS       2JL±1=  fa  +  4 

11  3  4  3  12 

Multiplying  by  12,  the  L.  C.  M.  of  3,  4,  12, 
120  +  6)  _  4(2a;  -  18)  4-  B(2x  +  3)  =  16  x  4  +  3a;  4-  4 
or  12(s  +  6)  _  8x  +  72  +  Gx  +  9  =  64  +  Bx  +  4> 

Transposing  and  reducing,  we  have 

12(*  +  6>  =  5x  -  13. 
11 

Multiplying  by  11,  we  have 

12(a;  +  6)  =  11  (5a;  -  13), 

or  12a;  +  72  =  55a:  -  143  ; 

.-.     43a;  =  215;         .-.     x  =  5. 

"We  may  verify  this  result  as  before. 

Note  3.  —  When  two  or  more  fractions  have  the  same  denominator, 
they  should  be  taken  together  and  simplified. 


EXAMPLES.  159 


4.    Solve  13~2a;  +  23a;  +  8i=16-^- 

x  +  3  4a  +  5  x  +  3 

Transposing,  we  have 

23a;  +  8^  _  4  =  16  -  *a;  -  13  +  2a?. 


4a;  +  5  x  +  3 

then  7a?  ~  ¥  =  3+fr 

4a;  +  5  a;  +  3 

Multiplying  across,  we  have 

(x  +  3)  (7a; 

or     7a;2  -  -%5-x  +  21a;  -  35  =  12a;  +  7a;2  +  15  +  -\5-x ; 

...     -l&x  =  50;         .-.     a;  =  -ffs. 

5.    Solve    g-8   +  ^=^  =  ^^  4-  ^^. 
35—10       a?  —  6       a;  —  7       x  —  9 

Note  4.  —  This  equation  might  be  solved  by  clearing  of  fractions, 
by  multiplying  by  the  four  denominators,  but  the  work  would  be  very 
laborious.  The  solution  will  be  much  simplified  by  transposing  two 
of  the  fractions  as  follows : 

Transposing,  we  have 

x  —  8   _  x  —  5  _  x  —  7       x  —  4 
x  —  10       x  —  7  ~~  a;  —  9       a?  —  6* 
Simplifying  each  side  separately,  we  have 
(a?-8)  (a;-7)-(a;-5)  (a;-10)  =  (x-7)  (x-(>)-(x-4)  (x-9) 

(ar-10)  (a;-7)  (x-9)  (aj-6) 

or 

a?-  15a;  +  56  -  (x2-  15a;  +  50)  =  a:2-13a;+42-(a;2-13a;+36) 
(a;-10)(a.--7)  (a;-9)  (a;-6) 

5 =  5 . 

(a?-10)  (a;-7)       (a?-9)(a;-6) 

Dividing  by  6  and  clearing  of  fractions,  we  have 

(a;  -  9)  (a;  -  6)  =  (x  -  10)  (a;  -  7), 

or  a;2  -  15a;  4-  54  =  a;2  -  17a;  +  70 ; 

.-.     x  =  8. 


160  EXAMPLES. 

Note  5.  —  This  example  may  also  be  solved  very  neatly  by  writing 
the  equation  at  first  in  the  form 

x  —  10  -f  2      x  —  6  +  2  _  x  — 7  +  2      x  — 0  +  2 

x  —  10  x  —  0  x  —  7  x  —  U    ' 

Reducing  each  fraction  to  a  mixed  number  (Art.  81),  we  have 

1  •  1    •  1,1  1,1 

which  gives  —  H = -\ 

3  x  —  10     x  —  6      a;  —  7     x  —  9 

Transposing, =  — — -. 

1        a      x  —  10      x  —  7      x  —  9      x  —  6 

3  3 


(x  -  10)  (x  -  7)       (x  -  9)(x  -  6)' 
and  the  solution  may  be  completed  as  before. 

„      c  ,       5a;  —  64       2a;  —  11       4a;  -  55       a?  —  6 
6.    Solve = . 

x  —  13  x  —  6  x  —  14        x  —  7 

Proceeding  as  in  the  second  method  of  Ex.  5,  we  have 

x  - 13        \         x-6j  x  - 14       \        x-l) 

1111 


x  —  13       x  —  6       a;  —  14       a:  — 
Simplifying  each  side  separately,  we  have 

7  7 


(a;  -  13)  (a;  -  6)        (a;  -  14)  (a:  -  7) 
Clearing  of  fractions,  or,  since  the  numerators  are  equal, 
the  denominators  must  be  equal,  we  have 

(a;  -  13)  (a;  -  6)  =  (x  -  14)  (x  -  7); 
.-.     x2  -  19x  +  78  =  x2  -  21a;  +  98  ; 
.-.     x  =  10. 
Solve  the  following  equations : 

Ans.  10. 


6. 


7. 
8. 
9. 

12.      1 
x        12a;       ^4* 
45                57 

2a;  +  3       4a;  —  5 
3a;  -  1        2a;  -  5 

x^-J  _  x  =  x  _? 

4  6 


HARDER  PROBLEMS.  161 


10.  6-^-±-8  -  2x  +  38  =  1.  Ans.  2. 
2x  +1         «  +  12 

11.  i(2»-10)-T1r(3a;-40)  =  15-i(57-.i-).    (See  Note 2). 

^4hs.  17. 


12. 

x—  1     a— 5     15  —  2a;_9  — a; 
4          32           40     "    2 

|5? 

aj  +  4         a;  +  5 

(See  Note  2).        5. 


14. 


3a;  - 

8 

6x  + 

13 

15 

3x  — 

1 

2x  - 

1 

6a;  + 

7 

3a;  +  5         2a; 
5a;  -  25  ~~   5  ' 


17. 


9x  +  6        x-        12a;  +  8 
4  2  5 


a;+3       a;+l       2a;+6      2a;+2 


(See  Note  1). 

20. 

*• 

-6f. 

(.     (See  Note  3). 

1. 

18.    ^^2  _|_  ® £  =  5 *  +  5 ^   (See  Ex.  5) .      4. 

a;  -  2       a;  -  6       a;  -  5       a;  -  3     v  ' 

,n     a;  —  1        a;  —  2        a;  —  5       a;  —  6  .. 

iy. = .  4*. 

a;  —  2        a;  —  3        a;  -6       a;—  7  J 

20     6a;  +  1  _    2-^  -  4   _  2a;  -  1  _2 
15            7a;  -  16             5 


92.  Harder  Problems  Leading  to  Simple  Equations 
with  One  Unknown  Quantity.  —  We  shall  now  give 
some  examples  which  lead  to  simple  equations,  but  which 
differ  from  those  of  Art.  61  in  being  rather  more  difficult. 
The  statement  of  the  problem  is  rather  more  difficult  than  in 
the  examples  of  that  Article,  and  the  equations  often  involve 
more  complicated  expressions. 


162  EXAMPLES. 


EXAMPLES. 


1.  A  alone  can  do  a  piece  of  work  in  9  days,  and  B  alone 
can  do  it  in  12  days:  in  what  time  will  they  do  it  if  they 
work  together? 

Let    x  =  the  number  of  days  required  for  both  to  do  the 
work  ; 

then       -  =  the  part  that  both  can  do  in  one  day. 

Also  i  =  the  part  that  A  can  do  in  one  day, 
and       i1^  =  the  part  that  B  can  do  in  one  day.. 

Since  the  sum  of  the  parts  that  A  and  B  separately  can 
do  in  one  day  is  equal  to  the  part  that  both  together  can  do 
in  one  day,  we  have 

*  +  «4 

Clearing  of  fractions  by  multiplying  by  36a;,  we  have 
Ax  +  3x  =  36;         .-.     x  =  5£, 
which  is  the  number  of  days  required. 

2.  A  workman  was  employed  for  60  days,  on  condition 
that  for  every  day  he  worked  he  should  receive  $3,  and  for 
every  day  he  was  absent  he  should  forfeit  $1  ;  at  the  end  of 
the  time  he  had  $48  to  receive :  required  the  number  of  days 
he  worked. 

Let  x  =  the  number  of  days  he  worked  ; 

then  60  —  x  =  the  number  of  days  he  was  absent. 
Also      3x  =  the  number  of  dollars  he  received, 
and   60  —  x  =  the  number  of  dollars  he  forfeited. 
Hence,  from  the  conditions  of  the  problem,  we  have 
3x  -  (60  -  x)  =  48. 

.-.     Ax  =  108.         .-.     x  =  27. 
That  is,  he  worked  27  days  and  was  absent  33  days. 

3.  A  starts  from  a  certain  place,  and  travels  at  the  rate 
of  7  miles  in  5  hours  ;  B  starts  from  the  same  place  8  hours 
after  A,  and  travels  in  the  same  direction  at  the  rate  of  5 


EXAMPLES.  163 

miles  in  3  hours  ;  how  far  will  A  travel  before  he  is  over- 
taken by  B? 

Let  x  =  the  number  of  hours  A  travels  before  he 

is  overtaken  ; 
then         x  —  8  =  the  number  of  hours  B  travels  before  he 
overtakes  A. 
Also  -|  =  the  part  of  a  mile  which  A  travels  in  one 

hour, 
and  -§  =  the  part  of  a  mile  which  B  travels  in  one 

hour, 
Therefore  %x  =  the  number  of  miles  which  A  travels  in  x 
hours, 
and     ^{x  —  8)  =  the  number  of  miles  which  B  travels  in 
x  —  8  hours. 
Since,  when  B  overtakes  A,  they  have  traveled  the  same 
number  of  miles,  we  have  for  the  equation 

f*  =  f  (x  -  8). 
.-.     21oj  =  25a;  -  200.         .-.     x  =  50. 

Therefore  f#  =  f  x  50  =  70  miles,  the  distance  which  A 
travels  before  he  is  overtaken  by  B. 

4.  A  cistern  could  be  filled  with  water  by  means  of  one 
pipe  alone  in  6  hours,  and  by  means  of  another  pipe  alone 
in '8  hours  ;  and  it  could  be  emptied  by  a  tap  in  12  hours  if 
the  two  pipes  were  closed :  in  what  time  will  the  cistern  be 
filled  if  the  pipes  and  the  tap  are  all  open? 

Let        x  =  the  required  number  of  hours. 
Then     £  =  the  part  of  the  cistern  the  first  pipe  fills  in 
one  hour ; 
therefore  -  =  the  part  of  the  cistern  the  first  pipe  fills  in 
x  hours. 
And       |  =  the  part  of  the  cistern  the  second  pipe  fills 
in  one  hour ; 
therefore  -  =  the  part  of  the  cistern  the  second  pipe  fills 
in  x  hours. 


164  EXAMPLES. 

Also 

hour ; 
therefore  —  =  the  part  of  the  cistern  the  tap  empties  in  x 
hours. 
Since  in  x  hours  the  whole  cistern  is  filled,  we  have,  repre- 
senting the  whole  by  unity, 

X       x x_  _  1 

6       8       12  ~    ' 
Multiplying  by  24,  we  have 

4*  +  Sx  -  2x  =  24. 
.-.  x  —  44- . 
5.  A  smuggler  had  a  quantity  of  brandy  which  he  ex- 
pected would  bring  him  $198 ;  after  he  had  sold  10  gallons 
a  revenue  officer  seized  one-third  of  the  remainder,  in  con- 
sequence of  which  the  smuggler  gets  only  $162  :  required 
the  number  of  gallons  he  had  at  first,  and  the  price  per 
gallon. 

Let  x  =  the  number  of  gallons  ; 

198 
then  ■ —  =  price  per  gallon  in  dollars. 

x  -  10 

o        =  the  number  of  gallons  seized  ; 


10  .,  198 


X  —  =  the  value  of   the  quantity  seized  . 


in 


3  it- 

dollars. 
Hence  we  have  the  equation 

«L=J°  x  128  _  m  _  162  =  86_ 

3  x 

Clearing  of  fractions 

66(3  —  10)  =  36a?. 

.-.     30aj  =  660. 

x  =  22,  the  number  of  gallons ; 

198        198        ftQ   .. 

and  =  — -  =  $9,  the  price  per  gallon. 

x  22 


EXAMPLES.  165 

6.  A  colonel,  on  attempting  to  draw  up  his  regiment  in 
the  form  of  a  solid  square,  finds  that  he  has  31  men  over, 
and  that  he  would  require  24  men  more  in  his  regiment  in 
order  to  increase  the  side  of  the  square  by  one  man  :  how 
many  men  were  there  in  the  regiment? 

Let  x  =  the  number  of   men   in   the   side 

of  the  first  square  ; 
then  or  -f  31  =  the  number  of  men  in  the  regiment. 

Also  (x  +  l)2  —  24  =  the  number  of  men  in  the  regiment. 
Hence,  we  have  the  equation 

x2  +  31  =  (x  +  l)2  -  24, 
or  x2  +  31  =  x2  +  2x  -  23.         .-.     x  =  27. 

Hence  (27)2  +  31  =  760  is  the  number  of  men  in  the 
regiment. 

Note  1.  — In  this  example  it  was  convenient  to  let  x  represent  the 
number  of  men  in  the  side  of  the  first  square  instead  of  the  number 
of  men  in  the  whole  regiment. 

7.  At  the  same  time  that  the  up-train  going  at  the  rate  of 
33  miles  an  hour  passes  A,  the  down-train  going  at  the  rate 
of  21  miles  an  hour  passes  B  ;  they  collide  18  miles  beyond 
the  midway  station  from  A  :  how  far  is  A*  from  B? 

Let  x  =  the  distance  from  A  to  B  in  miles  ; 

then  ®  =  half  the  distance. 

Also  -  +  18  =  the  number  of  miles  the  up-train  goes, 


and 

2 

At  _      distance  in  miles  ,,     ,.       .    , 

JNow : =  the  time  in  hours. 

rate  in  miles  per  hour 

^  +  18 
Therefore =  the  time  the  up-train  takes, 

DO 

-  -  18 
2 

and  — — —  =  the  time  the  down-train  takes. 

21 


166  EXAMPLES. 

Hence,  since  these  times  are  equal,  we  have  the  equation 
|  +  18        ~  -  18 


33  21 

Solving,  we  get  x  =  102,  which  is  the  distance  from  A  to 
B  in  miles. 

8.  A  cask,  A,  contains  12  gallons  of  wine  and  18  gallons 
of  water ;  and  another  cask,  B,  contains  9  gallons  of  wine 
and  3  gallons  of  water :  how  many  gallons  must  be  drawn 
from  each  cask  so  as  to  produce  by  their  mixture  7  gallons 
of  wine  and  7  gallons  of  water  ? 

Let  x  =  the  number  of  gallons  to  be  drawn  from  A  ; 

then  14  —  x  =  the  number  of  gallons  to  be  drawn  from  B, 
since  the  mixture  is  to  contain  14  gallons. 

Now  A -contains  30  gallons,  of  which  12  are  wine  ;  that  is, 
^|  of  A  is  wine.  Also  B  contains  12  gallons,  of  which  9  are 
wine  ;  that  is,  T9^  of  B  is  wine. 

Hence        3$x  =  the  number  of  gallons  of  wine  in  the 
x  gallons  drawn  from  A  ; 
and  x%(14  —  x)  =  the  number  of  gallons  of  wine  iu  the 

14  —  x  gallons  drawn  from  B. 
Since  the  mixture  is  to  contain  seven  gallons  of  wine,  we 
have  i2>T  +  T9_(14  -X)  =  7; 

that  is,  f«  -f     f  (14  -  x)  =  7. 

Solving,  we  get  as  =  10,  the  number  of  gallons  to  be  drawn 

from  A, 
and  14  —  x  =     4,  the  number  of  gallons  to  be  drawn 

from  B. 

9.  At  what  time  between  4  and  5  o'clock  is  the  minute- 
hand  of  a  watch  13  minutes  in  advance  of  the  hour-hand? 

Let  x  =  the  required  number  of  minutes  after  4  o'clock  ; 
that  is,  the  minute-hand  will  move  over  x  minute  divisions  of 
the  watch  face  in  x  minutes ;    and  as  it  moves  1 2  times  as 

x 
\2 
linute  divisions  in  x  minutes.    At  4  o'clock  the  minute-hand 


EXAMPLES.  167 

is  20  minute  divisions  behind  the  hour-hand,  and  finally  the 
minute-hand  is  13  minute  divisions  in  advance  ;  therefore,  in 
the  x  minutes,  the  minute-hand  moves  20  -f-  13,  or  33, 
divisions  more  than  the  hour-hand. 

Hence  a,*  =  -1-  +  33  ; 

therefore         lis  =12  x  33.         .-.     x  =  36, 
or  the  time  is  36  minutes  past  4. 

If  the  question  be  asked,  "  At  what  times  between  4  and 
o  o'clock  will  there  be  13  minutes  between  the  two  hands?  " 
we  must  also  take  into  consideration  the  case  when  the 
minute-hand  is  13  divisions  behind  the  hour-hand.  In  this 
case  the  minute-hand  gains  20  —  13,  or  7  divisions. 

Hence  x  =  —  +  7. 

.-.     11a;  =  84.         .-.     x  =  7TV 
Therefore   the   times   are    7^   minutes   past   4,    and    36 
minutes  past  4. 

Note  2.  -«-  The  student  is  supposed  to  have  obtained  from  Arith- 
metic some  knowledge  of  ratio  and  proportion.  When  two  or  more 
unknown  quantities,  in  any  example,  have  to  each  other  a  given  ratio, 
it  is  best  to  assume  each  of  them  a  multiple  of  some  other  unknown 
quantity,  so  that  they  shall  have  to  each  other  the  given  ratio.  Thus, 
if  two  unknown  numbers  are  to  each  other  as  2  to  3,  it  is  best  to 
express  the  numbers  by  2x  and  3a1,  since  these  two  numbers  are  to  each 
other  as  2  to  3.     This  will  be  illustrated  in  the  next  two  examples. 

10.  A  number  consists  of  two  digits  of  which  the  left 
digit  is  to  the  right  digit  as  2  to  3  ;  if  18  be  added  to  the 
number  the  digits  are  reversed  :  what  is  the  number? 

The  student  must  remember  that  any  number  consisting 
of  two  places  of  figures  is  equal  to  ten  times  the  figure  in 
the  ten's  place  plus  the  figure  in  the  unit's  place  ;  thus,  46  is 
equal  to  10x4  +  6;  likewise  358  is  equal  to  100  X  3 
+  10  X  5  +  8. 

Let  2x  =  the  left  digit ; 

then  3x  =  the  right  digit, 

and  10  x  2x  +  2>x  =  number. 


168  EXAMPLES. 

Hcuce  we  have  the  equation 

(10  x  2x  +  3a?)  +  18  =  (10  x  3x  +  2x), 
or  20a;  +  3a?  +  18  =  30a:  +  2a\ 

x  =  2 ; 
.-.     2a;  =  4,  and  3a:  =  G; 
therefore  the  number  is  4G. 

11.  A  hare  takes  4  leaps  to  a  greyhound's  3,  but  2  of  the 
greyhound's  leaps  are  equivalent  to  3  of  the  hare's  ;  the  hare 
has  a  start  of  50  leaps  :  how  many  leaps  must  the  greyhound 
take  to  catch  the  hare? 

Let  3.r  =  the  number  of  leaps  taken  by  the  greyhound  ; 
then     4a:  =  the  number  of  leaps  taken  by  the  hare  in  the 

same  time. 
Also,  let  a  denote  the  number  of  feet  in  one  leap  of  the  hare  ; 
then        fa  denotes  the  number  of   feet  in  one  leap  of  the 

greyhound. 
Therefore  3x  x  fa  =  the  distance  in  3a;  leaps  of  the  grey- 
hound ; 
and       (ix  +  50)  a  =  the  distance  in  Ax  +  50  leaps  of  the 

hare. 
Hence  we  have  the  equation 

fax-  =  (4a:  +  50)a. 
Dividing  by  a  and  multiplying  by  2,  we  have 

9a;  =  8a:  +  100.         .-.     x  =  100. 
Therefore  the  greyhound  must  take  300  leaps. 

Notk  3.  —  It  is  often  convenient  to  introduce  an  auxiliary  symbol, 
as  a  was  introduced  in  the  above  example,  to  enable  us  to  form  the 
equation  easily;  this  can  be  removed  by  division  when  the  equation  is 
formed. 

12.  A  person  bought  a  carriage,  horse,  and  harness  for 
$600  ;  the  horse  cost  twice  as  much  as  the  harness,  and  the 
carriage  half  as  much  again  as  the  horse  and  harness  :  what 
did  he  give  for  each?  Ans.  $360,  $160,  $80. 

13.  In  a  garrison  of  2744  men,  there  are  two  cavalry 
soldiers  to  twenty-five  infantry,  and  half  as  many  artillery 
as  cavalry:   lind  the  numbers  of  each.      Ans.  2150,  196,  98. 


EXAMPLES.  169 

14.  A  and  B  play  for  a  stake  of  $5  ;  if  A  loses  he  will 
have  as  much  as  B,  but  if  A  wins  he  will  have  three  times 
as  much  as  B  :  how  much  has  each?  Ans.  $25,  $15. 

15.  A,  B,  and  C  have  a  certain  sum  between  them ;  A 
has  one-half  of  the  whole,  B  has  one-third  of  the  whole,  and 
C  has  $50  ;  how  much  have  A  and  B?  Ans.  $150,  $100. 

16.  A  number  of  troops  being  formed  into  a  solid  square, 
it  was  found  that  there  were  60  over  j  but  when  formed  into 
a  column  with  5  men  more  in  front  than  before  and  3  less  in 
depth,  there  was  just  one  man  wanting  to  complete  it:  find 
the  number  of  men.  Ans.  1504. 

17.  A  and  B  began  to  pay  their  debts  ;  A's  money  was  at 
first  f  of  B's  ;  but  after  A  had  paid  $5  less  than  f  of  his 
money,  and  B  had  paid  $5  more  than  f  of  his,  it  was  found 
that  B  had  only  half  as  much  as  A  had  left :  what  sum  had 
each  at  first?  Ans.  $360,  $540. 

18.  In  a  mixture  of  copper,  lead,  and  tin,  the  copper  was 
5  lbs.  less  than  half  the  whole  quantity,  and  the  lead  and 
tin  each  5  lbs.  more  than  a  third  of  the  remainder :  find  the 
respective  quantities.  Ans.  20,  15,  15  lbs. 

19.  A  and  B  have  the  same  income  ;  A  lays  by  a  fifth  of 
his ;  but  B,  by  spending  annually  $400  more  than  A,  at  the 
end  of  four  years  finds  himself  $1100  in  debt:  what  was 
their  income?  Ans.  $625. 

20.  There  are  two  silver  cups  and  one  cover  for  both  ;  the 
first  weighs  12  ozs.,  and  with  the  cover  weighs  twice  as  much 
as  the  other  cup  without  it ;  but  the  second  with  the  cover 
weighs  a  third  as  much  again  as  the  first  without  it :  find  the 
weight  of  the  cover.  Ans.  6|  oz. 

21.  Two  casks,  A  and  B,  contain  mixtures  of  wine  and 
water ;  in  A  the  quantity  of  wine  is  to  the  quantity  of  water 
as  4  to  3  ;  in  B  the  like  proportion  is  that  of  2  to  3.  If  A 
contain  84  gallons  what  must  B  contain,  so  that  when  the 
two  are  put  together,  the  new  mixture  may  be  half  wine  and 
half  water?  Ans.  60. 


170  ""  EXAMPLES. 


EXAMPLES. 

Solve  the  following  equations. 

1.  2LzJ  +  4  =  x  -  2x  ~  1.  Ans.  -6. 

4  3 

2.  4*  +  17       to  -  10  =  ?j  2> 

a;  +  3  cc  —  4 

3.  ^-=-i  +  (a;  -  1)'(*  -  2)  =  x-  -  2x  -  4.  7. 

3 

4     3(7  +  Gx)  _  35  +  4x  1 

2  +  9a?  9  +  2x  ' 

X       +  -i-  =  1.  2. 


.  x  -f  2       a;  +  6 

„    21-5   .     a-3         4.x-  -  3       ,  . 
5  2x  -  15  10  liy 


„     4  (a;  +  3)  _  8a;  +  37  _  7x  -  29 
9  18  5x  -  12' 

7  60  10* 


4       5a;  -  30       3a;  —  12       x  —  6 


6. 
-10. 


Q     3a;2  -  2a;  -  8        (7a?  -  2)  (3a;  -  G)  0 

5  35 

10.  *±1°-  t(te-4)+  (»«-«)(»»-»)  =».-A.        2. 

3  G 

11.  — L_  +  -1-  =  — ^— .  |*. 

2a;  -  3       x  -  2       8a?  +  2  " J 

12     ft  —  4  _  x  —  5  _  x  —  7  _  x  —  8 


13.    _£-  +  "-^  =  *±^  +  2_ZL_8.  4. 

a;  —  2       x  —  7       x  —  1        a;  —  G 

- .     3  -  2a?       2a;  -  5        -  4a;2  -  1 

14. —     =     1     —     ;.  —  1. 

1  —  2x       2x  —  7  7  —  IGa;  +  4ar 

15.    "zJ+gMJ-^zl,^— +!+>.  -23. 


EXAMPLES. 

171 

16. 

3                   30                  3                 5 

^lns. 

-4. 

4  -  2x       8(1  -  x)        2  -  x        2  -  2a;' 

17. 

30  -f  Gx       60  +  8a;  _  u    ,       48 
x  +  1           x  +  3                    a;  +  l" 

3. 

18. 

x           x  +  1        a:  -  8       a;  —  9 

4 

as  —  2       a;  —  1       a;  —  6       a;  —  7 

19. 

x  +  5       a;  —  6  _  a;  —  4       a;  —  15 
a;  +  4       a;  —  7       a;  —  5       a;  —  16 

6. 

20. 

aj  —  7        a;  —  9        a;  —  13       x  —  15 
a;  —  9       a;  —  11       x  —  15       x  —  17 

13. 

21. 

a; +  3       a;  +  6  _  a;  +  2       cc  +  5 
o;  +  6       a;  +  9       a; +  5       a;  +  8 

-7. 

22. 

a;  +  2       x  —  7       a;  +  3  _  a;  —  6 
a;            a;  —  5       a;  +  1        a;  —  4 

2. 

23. 

4a;  -  17       10a;  -  13  _  8a;  -  30       5a;  - 

x  —  4            2a;  —  3           2a;  —  7          a;  — 

4 
1 

H. 

24. 

5aj  _  8       6a;  -  44       10a:  -  8  _  x  -  8 

a;—  2'      a;—  7            a;—  1          a;  —  6 

4. 

25.  (a;+lXa;+2Xa;+3)  =  (a;-lXa!-2Xa;-3)+3(4a;-2Xa;+l). 

Ans.  3. 

26.  (a;-9)  (a;-7)  (a;-5)(a;-l)  =  (a;-2)(a;-4)(a;-6)(a;-10). 

Ans.  5i. 


27.    (8a;  -  3)2(a;  -  1)  =  (4a;  -  l)2(4a;  - 

-5). 

*■ 

28.    x"  ~  x  +  l  +  a'2  +  x  +  1  =  2a;. 
a;  -  1                 a;  +  1 

0. 

Q9     6a;  +  7       2a;  —  2  _  2a;  +  1 

3. 

15  7a;  —  6  5 

30.  .ox  —  2  =  .25a;  +  .2a;  —  1.  20. 

31.  .5a;  +  .6a;  —  .8  =  .75a;  +  .25.  3. 

32.  .15.,  +  -135*  ~  "225  =  ^  -  '09x  ~  -18.  5. 

.6                  .2  .9 

33      2x  -  3    =     -4a;  ~  -6  p 

.3a;  -  .4       .06a;  -  .07'  2' 


172 

EXAMPLES. 

34. 

.3.t  -  1        .5  +  1.2a; 

.bx  -  A         2x  -  .1 

* 

35. 

(.3.i--  2)(.3.r-  1) 

.2x  -  1                  *(-3x 

2)  =  ,4a: 

-  2.      20. 

36. 

a?(x  —  a)  +  b2(x  —  6)  =  o&aj. 

a  +  b. 

37     2a;  +  3a  _  2  (3a;  +  2a) 
x  +  a  3x  +  a 

ss.  f(|  +  i)  =  *(|-i> 

»•«— >-(tP),-K-S-       £ 

40.    x2  +  a(2a  -  a?)  -  —  =  (x  -  -Y  -f-  a2.         a  +  6. 


a)^*+|t)  =  4^-.rVKa-4a;)(2a  +  3x). 


42. 

1                 1              a  -  b 

2ab 

x  —  a       x  —  b       x-  —  ab 

a  +  b 

43. 

x  —  a       x  +  a          2ax 
a  —  b       a  +  b       a2  —  6s 

a2 

b  —  a 

44. 

a!  —  a           a;  —  a  —  1 

a  - 

-  b 

X 

-&-  1 

a;  —  a  —  1       jb  —  a  —  2       a;  —  6  —  1       a;  —  6  —  2 

Jws.  $(a  +  &  +  3). 

45.    (a-«)3(x+«-2&)  =  (.T-&)3(.i--2«+&).       |(a+&). 

.  P      3a&c    ,       crlr (2a  +  6)?/2a;  _  „  te  afr 

a  +  &      (a  +  6) 3        a(a  +  6)s  «  '       a  +  &' 

47.  A  person  wishing  to  sell  a  watch  by  lottery,  charges 
$1.20  each  for  the  tickets,  by  which  he  gains  $1G  ;  whereas, 
if  he  had  made  a  third  as  many  tickets  again  and  charged 
$1  each,  he  would  have  gained  one-fifth  as  many  dollars  as 
he  had  sold  tickets  :   what  was  the  value  of  the  watch? 

Ana.  $128. 


EXAMPLES. 


173 


48.  There  is  a  number  of  two  digits,  whose  difference  is 
2,  and  if  it  be  diminished  by  half  as  much  again  as  the  sum 
of  the  digits,  the  digits  will  be  reversed :  find  the  number. 

Ans.  75. 

49.  Find  a  number  of  3  digits,  each  greater  by  1  than 
that  which  follows  it,  so  that  its  excess  above  a  fourth  of  the 
number  formed  by  reversing  the  digits  shall  be  36  times 
the  sum  of  the  digits.  Ans.  654. 

50.  A  can  do  a  piece  of  work  in  10  days,  which  B  can  do  in 
8  ;  after  A  has  been  at  work  upon  it  3  days,  B  comes  to  help 
him  :  in  how  many  days  will  they  finish  it?        Ans.  3^  days. 

51.  A  and  B  can  reap  a  field  together  in  7  days,  which  A 
alone  could  reap  in  10  days :  in  what  time  could  B  alone 
reap  it?  Ans.  23^  days. 

52.  A  privateer,  running  at  the  rate  of  10  miles  an  hour, 
discovers  a  ship  18  miles  off,  running  at  the  rate  of  8  miles 
an  hour :  how  many  miles  can  the  ship  run  before  it  is  over- 
taken^ Ans.  72. 

53.  The  distance  between  London  and  Edinburgh  is  360 
miles  ;  one  traveler  starts  from  Edinburgh  and  travels  at  the 
rate  of  30  miles  an  hour,  while  another  starts  at  the  same  time 
from  London  and  travels  at  the  rate  of  24  miles  an  hour : 
how  far  from  Edinburgh  will  they  meet?         Ans.  200  miles. 

54.  Find  two  numbers  whose  difference  is  4,  and  the  dif- 
ference of  their  squares  112.  Ans.  12,  16. 

55.  Divide  the  number  48  into  two  parts  so  that  the  excess 
of  one  part  over  20  may  be  three  times  the  excess  of  20  over 
the  other  part.  Ans.  32,  16. 

56.  A  cistern  could  be  filled  in  12  minutes  by  two  pipes 
which  run  into  it,  and  it  could  be  filled  in  20  minutes  by  one 
alone :  in  what  time  would  it  be  filled  by  the  other  alone  ? 

Ans.  30  minutes. 

57.  Divide  the  number  90  into  four  parts  so  that  the  first 
increased  by  2,  the  second  diminished  by  2,  the  third  multi- 
plied by  2,  and  the  fourth  divided  by  2,  may  all  be  equal. 

Ans.  18,  22,  10,  40. 


174  EXAMPLES. 

58.  Divide  the  number  88  iuto  four  parts  so  that  the  first 
increased  by  2,  the  second  diminished  by  3,  the  third  multi- 
plied by  4,  and  the  fourth  divided  by  5,  may  all  be  equal. 

Arts.  10,  15,  3,  60. 

59.  If  20  men,  40  women,  and  50  children  receive  $500 
among  them  for  a  week's  work,  and  2  men  receive  as  much 
as  3  women  or  5  children,  what  does  each  woman  receive  for 
a  week's  work?  Ans.  $5. 

GO.  A  cisterD  can  be  filled  in  15  minutes  by  two  pipes,  A 

^  and  B,  ruuuing  together ;  after  A  has  beeu  running  by  itself 

N^or  5  minutes  B  is  also  turned  on,  and  the  cistern  is  filled  in 

-^13  minutes  more :  in  what  time  would  it  be  filled  by  each 

pipe  separately?  Ans.  37|,  and  25  minutes. 

61.  A  man  and  his  wife  could  drink  a  cask  of  beer  in  20 
days,  the  man  drinking  half  as  much  again  as  his  wife  ;  but 
-§§  of  a  gallon  having  leaked  away,  they  found  that  it  only 
lasted  them  together  for  18  days,  and  the  wife  herself  for 
two  days  longer :  how  much  did  the  cask  contain  when  full  ? 

Ans.  12  gallons. 
Let  x  =  the  number  of  gallons  the  woman  could  drink  in  a  day. 

62.  A  man,  woman,  and  child  could  reap  a  field  in  30 
hours,  the  man  doing  half  as  much  again  as  the  woman,  and 
the  woman  two-thirds  as  much  again  as  the  child  :  how  many 
hours  would  they  each  take  to  do  it  separately  ? 

Ans.  62,  93,  155. 

Let  2x  =  the  man's  number  of  hours,  ox  =  the  woman's,  and  5x  = 
the  child's. 

63.  A  and  B  can  reap  a  field  together  in  12  hours,  A  and 
C  in  16  hours,  and  A  by  himself  in  20  hours  :  in  what  time 
(1)  could  B  and  C  together  reap  it,  and  (2)  could  A,  B,  and 
C  together  reap  it?  Ans.  21r9T  hours,  10^-!}  hours. 

64.  A  can  do  half  as  much  work  as  B,  B  can  do  half  as 
much  work  as  (',  and  together  they  can  complete  a  piece  of 
work  in  24  days  :  in  what  time  could  each  alone  complete 
the  work?  Ans.  168,  84,  and  42  days. 


EXAMPLES.  175 

65.  There  are  two  places  154  miles  apart,  from  which  two 
persons  start  at  the  same  time  with  a  design  to  meet ;  one 
travels  at  the  rate  of  3  miles  in  two  hours,  and  the  other  at 
the  rate  of  5  miles  in  four  hours :  when  will  they  meet  ? 

Ans.  At  the  end  of  56  hours. 

66.  Three  persons,  A,  B,  and  C,  can  together  complete  a 
piece  of  work  in  60  days  ;  and  it  is  found  that  A  does  three- 
fourths  of  what  B  does,  and  B  four-fifths  of  what  C  does  : 
in  what  time  could  each  one  alone  complete  the  work? 

Ans.  240,  180,  144  days. 
Let  x  =  C's  time  of  completing  the  work,  in  days. 

67.  A  general,  on  attempting  to  draw  up  his  army  in  the 
form  of  a  solid  square,  finds  that  he  has  60  men  over,  and 
that  he  would  require  41  men  more  in  his  army  in  order  to 
increase  the  side  of  the  square  by  one  man  :  how  many  men 
were  there  in  the  army?  Ans.  2560. 

68.  A  person  bought  a  certain  number  of  eggs,  half  of 
them  at  2  for  a  cent,  and  half  of  them  at  3  for  a  cent ;  he 
sold  them  again  at  the  rate  of  5  for  two  cents,  and  lost  a 
cent  by  the  bargain  :  what  was  the  number  of  eggs  ?    Ans.  60. 

69.  A  and  B  are  at  present  of  the  same  age ;  if  A's  age 
be  increased  by  36  years,  and  B's  by  52  years,  their  ages 
will  be  as  3  to  4  ;  what  is  the  present  age  of  each?   Ans.  12. 

70.  A  cistern  has  two  supply  pipes  which  will  singly  fill  it 
in  4£  hours  and  6  hours  respectively ;  and  it  has  also  a  leak 
by  which  it  would  be  emptied  in  5  hours  :  in  how  many  hours 
will  it  be  filled  when  all  are  working  together?         Ans.  5^. 

71.  A  person  hired  a  laborer  to  do  a  certain  work  on  the 
agreement  that  for  every  day  he  worked  he  should  receive  $2, 
but  that  for  every  day  he  was  absent  he  should  lose  $0.75  ;  he 
worked  twice  as  many  days  as  he  was  absent,  and  on  the  whole 
received  $39  :  how  many  days  did  he  work  ?  Ans.  24. 

72.  A  sum  of  money  was  divided  between  A  and  B,  so 
that  the  share  of  A  was  to  that  of  B  as  5  to  3  ;  also  the 
share  of  A  exceeded  five-ninths  of  the  whole  sum  by  $200  : 
what  was  the  share  of  each  person?  Ans.  $1800,  1080 


176  EXAMPLES. 

73.  A  gentleman  left  his  whole  estate  among  his  four  sons. 
The  share  of  the  eldest  was  $4000  less  than  half  of  the 
estate  ;  the  share  of  the  second  was  $600  more  than  one- 
fourth  of  the  estate  ;  the  third  had  half  as  much  as  the 
eldest ;  and  the  youngest  had  two-thirds  of  what  the  second 
had  :  how  much  did  each  son  receive  ? 

Ans.  $11000,  $8100,  $5500,  5400. 
Let  x  =  the  number  of  dollars  in  the  estate. 

74.  A  draper  bought  a  piece  of  cloth  at  38  cents  per  yard  ; 
he  sold  one-third  of  it  at  48  cents  per  yard,  one-fourth  of  it 
at  44  cents  per  yard,  and  the  remainder  at  40  cents  per  yard  ; 
and  his  gain  on  the  whole  was  $1.70  :  how  many  yards  did 
the  piece  contain?  Ans.  30. 

75.  A  and  B  shoot  by  turns  at  a  target ;  A  puts  7  bullets 
out  of  12  into  the  bull's  eye,  and  B  puts  in  9  out  of  12; 
between  them  they  put  in  32  bullets :  how  many  shots  did 
each  fire?  Ans.  24. 

76.  Two  casks,  A  and  B,  are  filled  with  two  kinds  of  sherry, 
mixed  in  the  cask  A  in  the  proportion  of  2  to  7,  and  in  the  cask 
B  in  the  proportion  of  2  to  5  :  what  quantity  must  be  taken 
from  each  to  form  a  mixture  which  shall  consist  of  2  gallons 
of  the  first  kind  and  6  of  the  second  kind  ?  Ans.  4£,  3|. 

77.  The  national  debt  of  a  country  was  increased  by  one- 
fourth  in  a  time  of  war.  During  a  long  peace  which  followed, 
$250000000  was  paid  off,  and  at  the  end  of  that  time  the 
rate  of  interest  was  reduced  from  \\  to  4  per  cent.  It  was 
then  found  that  the  amount  of  annual  interest  was  the  same 
as  before  the  war.  What  was  the  amount  of  the  debt  before 
the  war?  Ans.  $2000000000. 

78.  How  many  minutes  does  it  want  of  4  o'clock,  if  three- 
quarters  of  an  hour  ago  it  was  twice  as  many  minutes  past 
2  o'clock?  Ans.  25. 

Let  x  =  the  number  of  minutes  it  wants  of  4  o'clock. 

7!).  At  whal  time  between  3  o'clock  and  4  o'clock  is  one 
hand  of  a  watch  exactly  in  the  direction  of  the  other  hand 
produced?  Ans.  41)^  minutes  past  three. 


EXAMPLES.  177 

80.  The  hands  of  a  watch  are  at  right  angles  to  each  other 
at  3  o'clock  :  when  are  they  next  at  right  angles? 

Ans.  32T8T  minutes  past  three 

81.  At  what  time  between  3  and  4  o'clock  is  the  minute 
hand  one  minute  ahead  of  the  hour-hand? 

17^  minutes  past  three 

82.  A  clock  has  two  hands  turning  on  the  same  centre  ; 
the  swifter  makes  a  revolution  every  12  hours,  and  the  slower 
every  16  hours :  in  what  time  will  the  swifter  gain  just  one 
complete  revolution  on  the  slower?  Ans.  48  hours. 

83.  A  watch  gains  as  much  as  a  clock  loses  ;  and  1799 
hours  by  the  clock  are  equivalent  to  1801  hours  by  the 
watch :  find  how  much  the  watch  gains  and  the  clock  loses 
per  hour.  Ans.  2  seconds. 

Let  x  ■=  the  number  of  seconds  which  the  watch  gains  and  the 
clock  loses  per  hour. 

84.  A  and  B  made  a  joint  stock  of  $2000  by  which  they 
gained  $640,  of  which  A  had  for  his  share  $128  more  than 
B  :  what  did  each  contribute  to  the  stock  ? 

Ans.  $1200,  $800. 

85.  A  hare  is  80  of  her  own  leaps  ahead  of  a  greyhound  ; 
she  takes  3  leaps  for  every  2  that  he  takes  ;  but  he  covers  as 
much  ground  in  one  leap  as  she  does  in  2  :  how  many  leaps 
will  the  hare  have  taken  before  she  is  caught?         Ans.  240. 

86.  A  and  B  play  at  a  game,  agreeing  that  the  loser 
shall  always  pay  to  the  winner  one  dollar  less  than  half  the 
money  the  loser  has  ;  they  commence  with  equal  quantities 
of  money,  and  after  B  has  lost  the  first  game  and  won  the 
second,  he  has  $2  more  than  A :  how  much  had  each  at 
the  commencement?  Ans.  $6. 

87.  An  officer  can  form  his  men  into  a  hollow  square  4 
deep,  and  also  into  a  hollow  square  8  deep  ;  the  front  in  the 
latter  formation  contains  16  men  fewer  than  in  the  former 
formation  :  find  the  number  of  men.  Ans.  640. 

88.  A  person  has  just  a  hours  at  his  disposal ;  how  far 
may  he  ride  in  a  coach  which  travels  b  miles  an  hour,  so  as 


178  EXAMPLES. 


to  return  home  in  time,  walking  back  at  the  rate  of  c  miles 


an  hour?  Ans. 


abc 


b  +  c 

89.  A  man  rides  one-third  of  the  distance  from  A  to  B  at 
the  rate  of  a  miles  an  hour,  aud  the  remainder  at  the  rate  of 
26  miles  an  hour  ;  if  he  had  traveled  at  a  uniform  rate  of  3c 
miles  an  hour,  he  could  have  ridden  from  A  to  B  and  back 
again  in  the  same  time  :  prove  that 

-  =  l  +  l- 

cab 

90.  A  certain  article  of  consumption  is  subject  to  a  duty 
of  $2.25  per  cwt.  ;  in  consequence  of  a  reduction  in  the 
duty  the  consumption  increases  one-half,  but  the  revenue 
falls  one-third :  find  the  duty  per  cwt.  after  the  reduction. 

Ans.  SI. 

91.  A  vessel  can  be  emptied  by  three  taps;  by  the  first 
alone  it  could  be  emptied  in  80  minutes,  by  the  second  alone 
in  200  minutes,  and  by  the  third  alone  in  5  hours.  In  what 
time  will  the  vessel  be  emptied  if  all  the  taps  are  opened  ? 

Ans.  48  minutes. 

92.  A  cask  A  contains  12  gallons  of  wine  and  18  gallons 
of  water ;  and  another  cask  B  contains  9  gallons  of  wine  and 
3  gallons  of  water :  how  many  gallons  must  be  drawn  from 
each  cask  so  as  to  produce  by  their  mixture  7  gallons  of 
water  and  7  gallons  of  wine  ?        Ans.  10  from  A,  4  from  B. 

93.  A  ship  sails  with  a  supply  of  biscuit  for  60  days,  at  a 
daily  allowance  of  a  pound  a  head  ;  after  being  at  sea  20 
daj's  she  encounters  a  storm  in  which  5  men  are  washed 
overboard,  and  damage  sustained  that  will  cause  a  delay  of 
24  days,  and  it  is  found  that  each  man's  daily  allowance 
must  be  reduced  to  five  sevenths  of  a  pound.  Find  the  orig- 
inal number  of  the  crew.  Ans.  40. 


SIMULTANEOUS  EQUATIONS.  179 


CHAPTER    X. 

SIMULTANEOUS    SIMPLE    EQUATIONS    OF   TWO 
OR    MORE    UNKNOWN    QUANTITIES. 

93.  Simultaneous  Equations  of  Two  Unknown 
Quantities.  —  If  we  have  a  single  equation  containing  two 
unknown  quantities  x  and  y,  we  cannot  determine  any  thing 
definite  regarding  the  values  of  x  and  y,  because  whatever 
value  we  choose  to  give  to  either  of  them,  there  will  be  a 
corresponding  value  of  the  other. 

Thus,  from  the  equation, 

2x  +  Sy  =  24, (1) 

we  may  deduce  the  equation, 

24  -  2x 

v  = ; 

u  3 

but  we  cannot  find  the  value  of  y  from  this  equation  unless 
we  know  the  value  of  x.  "We  may  give  to  x  any  value  we 
choose,  and  there  will  be  one  corresponding  value  of  y ;  and 
thus  we  may  find  as  many  pairs  of  values  as  we  please  which 
will  satisfy  the  given  equation. 

For  example,  if  x  =    3,  then  y  =  (24 

If  x  =     6,  then  y  =  (24 

If  x  =     9,  then  y  =  (24 

If  x  =  20,  then  y  =  (24 
and  so  on. 

Any  one  of  these  pairs  of  values  f    ""'     },  (  ),  (  \ 

J  \y=«J  \y=y  \y=i) 

etc.,  substituted  for  x  and  y  in  (1)  will  satisfy  the  equation. 
Hence  a  single  equation  containing  two  unknown  quantities 
is  not  sufficient  to  determine  the  definite  value  of  either. 


6)- 

-3  =  6. 

12)- 

-3  =  4. 

18)- 

-3  =  2. 

40)  - 

-  3  =  -51 

180  SIMULTANEOUS  EQUATIONS. 

Suppose  we  have  a  second  equation  of  the  same  kind, 
expressing  a  different  relation  between  the  unknown  quanti- 
ties, as  for  example, 

3x  +  2y  =  26  ; (2) 

then  we  can  find  as  many  pairs  of  values  as  we  please  which 
will  satisfy  this  equation  also. 

Now  suppose  we  wish  to  determine  values  of  x  and  y 
which  will  satisfy  both  equations  (1)  and  (2)  ;  we  shall  find 
that  there  is  only  one  pair  of  values  of  x  and  y,  i.e.,  only 
one  value  of  x  and  one  value  of  y  that  will  satisfy  both 
equations.  For,  multiply  equation  (1)  by  2,  and  equation 
(2)  by  3,  and  the  equations  become 

4X  +  Qy  =  48, (3) 

and  9a;  +  Gy  =  78 (4) 

The  coefficients  of  y  are  now  the  same  in  (3)  and  (4)  ; 
hence  if  we  subtract  each  member  of  (.">)  from  the  corre- 
sponding member  of  (4),  we  shall  obtain  an  equation  which 
does  not  contain  y  :  the  equation  will  be 

5<b  =  30 ; 
therefore  x  =  0. 

Substituting  this  value  of  x  in  either  of  the  two  given 
equations,  for  example  in  equation  (1),  we  have 
12  +  3//  =  24. 
.-.     3y  =  12. 
y  =  4. 
Thus,  if  both  equations  are  to  be  satisfied  by  the  same 
values  of  x  and  y,  x  must  equal  G,  and  y  must  equal  4  ;  and 

the  pair  of  values  f  )  is  the  only  pair  of  values  which 

\y  =  v 

will  satisfy  both  the  given  equations. 

Simultaneous  Equations  are  those  which  are  satisfied  by 
the  same  values  of  the  unknown  quantities.     Thus, 

Since  (1)  and  (2)  are  satisfied  by  the  same  values  of  x 
and  ?/,  they  are  simultaneous  equations. 


ELIMINATION   BY  ADDITION   OR   SUBTRACTION.      181 

Independent  Equations  are  those  which  express  different 
relations  between  the  unknown  quantities,  and  neither  can 
be  reduced  to  the  other.     Thus, 

Equations  (1)  and  (2)  are  independent,  because  they  ex- 
press different  relations  between  x  and  y.  But  2x  —  Sy  =  4, 
and  8.1-  —  12y  =  16,  are  not  independent  equations,  since 
the  second  is  derived  directly  from  the  first,  by  multiplying 
both  members  by  4. 

Hence  we  see  that  two  independent  simultaneous  equations 
are  necessary  to  determine  the  values  of  two  unknown  quan- 
tities. 

94.  Elimination.  —  In  order  to  solve  any  two  simulta- 
neous equations  containing  two  unknown  quantities,  it  is 
necessary  to  combine  them  in  such  a  way  as  to  deduce  a 
third  equation  which  contains  only  one  of  the  unknown 
quantities  ;  and  this  equation  containing  only  one  unknown 
quantity  can  be  solved  by  the  method  given  in  Chapter  IX. 
When  the  value  of  one  of  the  unknown  quantities  has  thus 
been  determined,  we  can  substitute  this  value  in  either  of 
the  given  equations,  and  then  determine  the  value  of  the 
other  unknown  quantity. 

The  process  of  combining  equations  so  as  to  get  rid  of 
either  of  the  unknown  quantities  is  called  Elimination.  The 
unknown  quantity  which  disappears  is  said  to  be  eliminated. 

There  are  three  methods  of  elimination  in  common  use :  * 
(1)  by  Addition  or  Subtraction;  (2)  by  Substitution;  and 
(3)  by  Comparison. 

95.  Elimination  by   Addition   or   Subtraction.— 

1.    Let  it  be  required  to  determine  the  values  of  x  and  y  in 
the  two  equations 

8a;  +  1y  =  100 (1) 

12jb  -  %  =88 (2) 


*  There  is  also  a  method  by  Undetermined  Multipliers,  which  sometimes  lias  the 
advantage  over  either  of  these  three  methods,  especially  in  Higher  Mathematics. — 
See  Art.  99,  Note. 


182      ELIMINATION   BY   ADDITION   OR   SUBTRACTION. 

If  we  wish  to  eliminate  y  we  multiply  (1)  by  5,  and  (2) 
by  7,  so  as  to  make  the  coefficients  of  y  in  both  equations 
equal.     This  gives 

40a;  +  35y  =  500 (3) 

84b  -  S5y  =  GIG (4) 

Adding  (3)  and  (4),  we  have 

124x  =  1116.         .-.     x  =  9. 

To  find  y,  substitute  this  value  of  x  in  either  of  the  given 
equations.     Thus  in  (1) 

72  +  ly  =  100. 
.-.     ly  =     28. 

y  =     4 

and  x  =       9 

In  this  solution  we  eliminated  y  by  addition. 

Otherwise  thus:  Suppose  that  in  solving  these  equations 
we  wish  to  eliminate  x  instead  of  y.  Multiply  (1)  by  3,  and 
(2)  by  2,  so  as  to  make  the  coefficients  of  x  in  both  equations 
equal.     This  gives 

2ix  +  21y  =  300 (5) 

24a>  -  10?/  =  17G (6) 

Subtracting  (G)  from  (5),  we  have 

31y  =  124.         .-.     y  =  4. 

In  this  solution  we  eliminated  x  by  subtraction. 

Note  1.  —  The  student  will  observe  that  we  might  have  made  the 
coefficients  of  x  equal  by  multiplying  (1)  and  ('2)  by  12  and  8  respec- 
tively, instead  of  by  3  and  2;  but  it  was  more  convenient  to  use  the 
smaller  multipliers,  because  it  enabled  us  to  work  with  smaller 
numbers. 

2.    Solve  2a?  -f-    Sy  =  81 (1) 

12.»;  -  17y  =  -59 (2) 

Here  it  will  be  more  convenient  to  eliminate  x. 


ELIMINATION   BY   ADDITION   OR  SUBTRACTION.      183 

Multiplying  (1)  by  G,  to  make  the  coefficients  of  x  in  both 
equations  equal,  we  have 

12a;  +  I82/  =186;      ....  (3) 
and  from  (2)  12a;  -  Uy  =  -59  .....   (4) 

Subtracting  (4)  from  (3),  35y  =     245. 

■■■    y  =       7. 

Substituting  this  value  of  y  in  (1),  we  have 
2x  4-  21  =  31. 

.  • .     x  =     51 
and  y  =     7j " 

Note  2.  —  When  one  of  the  unknown  quantities  has  been  found,  it 
is  immaterial  which  of  the  equations  we  use  to  complete  the  solution, 
though  it  is  sometimes  more  convenient  to  use  a  particular  equation 
on  account  of  its  being  less  involved  than  the  other.  Thus,  in  this 
example,  we  substituted  the  value  of  x  in  (1)  rather  than  in  (2), 
because  it  rendered  the  process  simpler. 

In  these  two  examples  we  have  eliminated  by  addition  and 
subtraction.     Hence  to  eliminate  an  unknown  quantity  by 
addition  or  subtraction,  we  have  the  following 
Rule. 

Multiply  the  given  equations,  if  necessary,  by  such  numbers 
as  will  make  the  coefficients  of  this  unknown  quantity  numeri- 
cally equal  in  the  resulting  equations.  Then,  if  these  equal 
coefficients  have  unlike  signs,  add  the  equations  together;  if 
they  have  the  same  sign,  subtract  one  equation  from  the  other. 

Rem. —  It  is  generally  best  to  eliminate  tbat  unknown  quantity 
which  has  the  smaller  coefficients  in  the  two  equations,  or  which 
requires  the  smallest  multipliers  to  make  its  coefficients  equal.  When 
the  coefficients  of  the  quantity  to  be  eliminated  are  prime  to  each 
other,  we  may  take  each  one  as  the  multiplier  of  the  other  equation. 
When  these  coefficients  are  not  prime  to  each  other,  find  their  least 
common  multiple;  and  the  smallest  multiplier  for  each  equation  will 
be  the  quotient  obtained  by  dividing  this  L.  C.  M.  by  the  coefficient  in 
that  equation.  Thus,  in  Ex.  1,  first  solution,  7  and  5  (the  coefficients 
of  y)  are  prime  to  each  other.  We  multiplied  (1)  by  5  and  (2)  by  7. 
In  the  second  solution  of  Ex.  1,  the  L.  C.  M.  of  S  and  12  (the  coeffi- 
cients of  .r)  is  24;  and  hence  the  smallest  multipliers  of  (1)  and  (2) 
are  3  and  2  respectively,  which  we  used  in  that  solution. 


184      ELIMINATION   BY  ADDITION   OR   SUBTRACTION. 

3.  Solve  171a;  —  213?/  =  642 (1) 

111.-'  -  320?/  =244 (2) 

Here  we  see  that  171  and  114  contain  a  common  factor  57  ; 
so  we  shall  make  the  coefficients  of  x  in  (1)  and  (2)  equal 
to  the  least  common  multiple  of  171  and  114  if  we  multiply 
(1)  by  2  and  (2)  by  3. 
Thus,  342a  -  426#  =  1284 

342a  -  978y  =     732 
Subtracting,  552//  =     552. 

.-.    y=  i,l 

therefore  a-  =  5.  j 

Note  3.  — The  solution  is  sometimes  easily  effected  by  first  adding 
the  given  equations,  or  by  subtracting  one  from  the  other.     Thus, 

4.  Solve  127a  +    5%=  1928.     .     .     .   (1) 

59a;  +  I27y  =  171)2.  .  .  .  (2, 
By  addition                        186a  -4-  186y  =  3720. 

.-.  x  +  y  =  20  .  .  .  .  (3) 
Subtracting  (2)  from  (1),  68a;  -  68?/  =  136. 

•••     x-y  =2      ....    (4) 

Adding  (3)  and  (4),                          2a;  =  22.  .*.  x  =  \\. 

Subtracting  (4)  from  (3),                  2//  =  18.  .-.  y  =     9. 

Note  4. — The  student  should  look  carefully  for  opportunities  to 
effect  such  reductions  as  are  made  in  this  example.  He  will  find  as  he 
proceeds  that  in  all  parts  of  Algebra,  particular  examples  may  be 
treated  by  methods  which  are  shorter  than  the  general  rules;  but  such 
abbreviations  can  only  he  suggested  by  experience  and  practice. 

Solve  the  following  equations  by  addition  or  subtraction: 

5.  3a;  +  4y  =  10,  4a  +  ?/  =  9.  Ans.  x  =  2,y  =  1. 

6.  x  +  2y  =  13,  3a  +  ?/  =  It.  x  =  3,  y  =  5. 

7.  4a;  +  ly  =  29,  x  +  3y  =11.  x  =  2,  y  =  3. 

8.  8a  -  ?/  =  34,  a;  +  8?/  =  53.  x  =  5,  y  =  6. 

9.  14a;  -  3y  =  39,  6a  +  17?/  =  35.  x  =  8,  y  =  1. 
10.  35a  +  17#  =  86,  56a  -  Vdy  =17.         a  =  1,  y  =  3. 


ELIMINATION   BY  SUBSTITUTION.  185 

11.  15a;  +  lly  =  92,  55a;  —  33?/  =  22. 

Ans.  x  =     1,  y  =     1. 

12.  3a?  +  2y  =  32,  20a;  -  3y  =  1.  a:  =    2,  ?/  =  13. 

13.  7x  +  5y  =  60,  13a;  —  lly  t=  10.       a-  =    5,  ?/  =    5. 

14.  10s  -f-  9y  =  290,  12a;  -  lly  =  130.    a-  =  20,  y  =  10. 

96.  Elimination  by  Substitution.  —  Find  the  values 
of  x  and  y  in  the  equations 

4a; '.+  Sy  =  22 (1) 

5x  —  7y  =    G (2) 

Transpose  3?/  in  (1),  4a;  =  22  —  3?/; 

divide  by  4,  a;  =  22  ~  3y ; 

substitute  this  value  of  a;  in  (2),  and  we  obtain 
,/22  -  3?/\ 

5(— 4       j-7'/=     G; 
multiply  by  4,     5(22  -  %)  -  28y  =  24. 
.-.     ?/  =     2. 
Substitute  this  value  of  y  in  eitfier  (1)  or  (2),  thus  in  (1) 
4a;  +  G  =  22. 
.-.     x  =  4 
and  ?/  =  2  ' 

In  this  solution  we  eliminated  a;  by  substitution. 
Otherwise  thus:  from  (1)  we  have 
3?/  =  22  -  4a; ; 

divide  by  3,  y  =  £lz^; 

o 

substitute  this  value  of  y  in  (2), 

multiply  by  3,      15a;  -  7(22  -  Ax)  =  18  ; 

that  is,  15a;  —  154  +  28a;  =  18. 

.-.     x  =    4. 


1SG  ELIMINATION   BY   COMPARISON. 

Substitute  this  value  of  x  in  either  (1)  or  (2),  thus  in  (1) 

16  +  By  =  22.         .-.     y  =  2. 
Here  we  eliminated  y  by  substitution. 

Hence,  to  eliminate  an  unknown  quantity  by  sxibstitution, 
we  have  the  following 

Rule. 

From  either  equation,  find  the  value  of  the  unknown  quan- 
tity to  be  eliminated,  in  terms  of  the  other;  and  substitute  this 
value  for  that  quantity  in  the  other  equation. 

Solve  by  substitution  the  following  equations  : 

2.  3a;  —  \y  =  2,  Ix  —  9y  =  7.    Ans.  x  =  10,  y  =  7. 

3.  11a,*  —  ly  =  37,  8a;  -f  9y  =  41.        a;  =     4,  ?/  =  1. 

4.  Ga;  —  ly  =  42,  7x  —  Gy  =  75.          a?  ==  21,  ?/  =  12. 

5.  3a;  -  4?/  =  18,  3a;  +  2y  =  0.            a;  =     2,  ?/  =  -3. 

6.  4a;  -2=11,  2a;  -  By  =  0.  a>  =    3,  y  =      2. 

7.  2x  —  y  =  9,  3x  —  ly  =  19.  x  —    4,  ?/  =  —  1. 

8.  15a;  +  ly  =  29,  9a;  +  15?/  =  39.      a;  =     1,  y  =       2. 

9.  2aj  +  y  =  10,  7a;  -f  8?/  =  53.  a;  =     3,  ?/  =      4. 

97.  Elimination  by  Comparison.  —  Find  the  values 
of  x  and  y  in  the  equations 

2a?  +  By  a  23 (1) 

5a;  -  2y  =  10     .     .  '  .     .     .     .   (2) 

Finding  the  value  of  x  in  terms  of  y  from  both  (1)  and 
(2),  we  have, 
from(l),  x  =  23  ~  3?/, (3) 

and  from  (2),  a;  =  10  +  2y (1) 

5 

Placing  these  two  values  of  x  equal  to  each  other,  we  have 

10  +  2y       23  -  3?/ 

5  2 


ELIMINATION  BY   COMPARISON.  187 

Clearing  of  fractious,  by  multiplying  by  10,  we  have 
20  +  4?/  =  115  —  Voy. 
.-.     19?/  =     95.         .-.     y  =  5. 
Substitute  this  value  of  y  in  either  (3)  or  (4),  thus  in  (4) 

10  +  10 

x  = ! =  4. 

5 

In  this  solution  we  eliminated  x  by  comparison. 
Otherwise  thus:  find  the  values  of  y  in  terras  of  x  from 
(1)  and  (2). 

(5) 


Therefore 


2/ 

= 

23 

— 

2x 

3 

y 

= 

5x 

— 

10 

2 

2z 

5a 

- 

10 

J3  _- 

3  2 

Clearing  of  fractions 

46  -  4flj  =  15a?  -  30. 
.-.     19a;  =  76. 

x  —    4,  the  same  as  before. 
Substituting  this  value  of  x  in  either  (5)  or  (6),  we  deduce 
y  =  5,  as  before. 

In  this  solution  we  eliminated  y  by  comparison. 
Hence,  to  eliminate  an  unknown  quantity  by  comparison, 
we  have  the  following 

Rule. 

From  each  equation  find  the  value  of  the  unknown  quan- 
tity to  be  eliminated,  in  terms  of  the  other;  then  place  these 
values  equal  to  each  other. 

Note.  —  Either  of  these  methods  of  elimination  may  be  employed, 
according  to  circumstances,  and  we  shall  always  obtain  the  same 
result,  whichever  one  we  use;  each  method  has  its  advantages  in 
particular  cases.  Generally,  the  last  two  methods  introduce  fractional 
expressions,  while  the  first  method  does  not,  if  the  equations  be  first 
cleared  of  fractions.    As  a  general  rule,  the  method  by  addition  or 


188  FRACTIONAL  SIMULTANEOUS  EQUATIONS. 

subtraction  is  the  most  simple  and  elegant.  When  either  of  the  un- 
known quantities  has  1  for  its  coefficient,  the  method  by  substitution 
is  advantageous.  When  there  are  more  than  two  unknown  quantities, 
it  is  often  convenient  to  use  several  of  the  methods  in  the  same 
example. 

Solve  by  comparison  the  following  equations  : 

2.  Ix  —  by  =  24,  4a:  —  Sy  =  11.     Ans.  x  =  17,  y  =  19. 

3.  |  +  Sy  =  7,  ^L_pJ  =  32/  -  4.  x=    3,    y  =  2. 

4.  6*  -  by  =  1,  7«  —  4?/  =  8|.  a;  =    3£,  y  =  4. 

5.  1±J/  +  a  =  15,  ?L=J?  +y  =  6.      a;  =  10,    y  =  5. 

o  o 

6.  —  +  5w  =  13,  2x  +  4  ~  ly  =  33.       a:  =  19,    y  =  2. 
19         J  '2  '    * 

7.  2a  + t-?  =  21,4?/  +  ^i  =  29.       a;  =  10,    y  =  7. 

5  b 


98.    Fractional    Simultaneous    Equations    of   the 

Form  12  +  8    =  

x        y 

*1  -  H  =  3 (2) 

x         y 

If  we  cleared  these  equations  of  fractions  they  would 
involve  the  product  xy  of  the  unknown  quantities  ;  and  thus 
they  would  become  quite  complex.  But  they  may  be  solved 
by  the  methods  already  given,  as  follows  : 


Multiply  (1)  by  3, 
Multiply  (2)  by  2, 

3G 

X 

54 

X 

+  !*  =  24. 

y 

-M-   6. 

y 

Add 

Divide  by  30, 

?2=30. 

X 

5=   i. 

x  =  3. 


LITERAL   SIMULTANEOUS   EQUATIONS.  180 


Substitute  this  ^ 

'alue  of  x  in  (1) 

, 

Transpose, 

+  §  =  8. 

y 

*  =  4. 

y 

ring  equations  : 

y  =  2. 

Solve  the  follow 

2.  •_* 

x        y 

= 

!,  18  +  ?0  = 

16. 

^4ns. 

a;  =  3,  7/  =  2. 

3.  §_» 

a;        y 

= 

i,i5+  «  = 

a;         ?/ 

7. 

x  =  2,  y  =  3. 

4.    5+6 
a;        ?/ 

= 

3,  15  +  ?  _ 
*         2/ 

4. 

x  =  5,  ?/  =  3. 

5.  5_  2 

a;        ?/ 

= 

»         2/ 

3. 

a;  =  2,  ?/  =  7. 

6.   5+10 

*        2/ 

= 

79, 1«  -  1  = 

x         y 

44. 

*  =  h  y  =  i- 

99.    Literal   Simultaneous   Equations.  —  Let 

required  to  solve 

ax     +  by       =  c    .     .     . 

a'#  +  &'?y  =  c'  .  .  . 
Multiply  (1)  by  a',  aa'as  -+-  a'62/  =  a'c  •  •  • 
Multiply  (2)  by  a,  aa'a  +  ab'y    =  ac' .     .     . 

Subtract  (3)  from  (4),  (ab'    —  a'b)y  —  ac'  —  a'c. 
Divide  by  (ab'  —  a'b) ,  y  = 


it   be 


(1) 
(2) 
(3) 
(4) 


a&'  —  a'6 

To  find  the  value  of  x,  eliminate  y  from  (1)  and  (2)  thus : 
Multiply  (1)  by  &'  and  (2)  by  6, 

ab'x  +  bb'y  =  b'c    .     .     .     .   (5) 
a'&a;  +  bb'y  =  be'   .     .     .     .   (G) 

Subtract  (6)  from  (5),  (ab'  —  a'&)aj  =  &'c  —  be'. 

b'c  -  6c' 


Divide  by  (ab'  —  a'&), 


ab'  -  a'b 


190  LITERAL   SIMULTANEOUS  EQUATIONS. 

Note. — As  previously  explained  (Art.  95),  this  value  of  x  might 
be  found  by  substituting  the  value  of  y  in  either  (1)  or  (2);  but 
sometimes  the  value  of  the  second  unknown  quantity  is  more  easily 
found  by  elimination,  as  in  this  case,  than  by  substituting  the  value, 
of  the  unknown  quantity  already  found.  The  method  by  Undeter- 
mined Multipliers  is  often  advantageous  in  the  solution  of  literal 
equations. 

Method  by  Undetermined  Multipliers.  —  Let  it  be  required  to  solve 
by  undetermined  multipliers  the  same  equation  as  in  the  last  example. 

ax  +  by  =  c (1) 

a'x  +  b'y  =  c' (2) 

Multiply  either  (1)  or  (2)  by  any  finite  constant  whatever;  thus, 
multiply  (2)  by  m, 

ma'x  +  mb'y  =  mc' (3) 

Add  (1)  and  (3),  ax  +  by  +  ma'x  +  mb'y  =  c  +  mc'. 

Factoring,  (a  +  ma')x  +  (b  +  mb')y  =  c  +  mc'  ....   (4) 

Now  since  m  is  at  our  disposal,  we  may  so  choose  it  that  the 
coefficient  of  y  in  (4)  shall  be  zero;  that  is, 

b  +  mb'=  0; (5) 

and  (4)  will  reduce  to       (a  -f  ma')x  =  c  +  mc' (6) 

From  (5)  wc  have  m  =  — -^, 

and  substituting  this  value  for  m,  we  deduce  from  (6) 

_  c  +  mc'  _  b'  _  be  —  be' 

~  a  +  ma'         _  ba/      ab'  —a'b 
b' 
The  value  of  y  may  next  be  obtained  by  so  choosing  m  that  the 
coefficient  of  x  in  (4)  shall  be  zero;  that  is, 

a  +  ma'  =  0, • (7) 

(b  +  mb')y  —  c  +  mc'.    .......  (8) 

From  (7)  we  have  m  =  —  - \, 

a 

and  using  this  value  for  m,  we  deduce  from  (8) 

_  c  +  mc'  _     (£_  _  a'c  —  ac'. 

~  b  +  mb'  ~  ,  _  ab'      a'b  —  o&'' 
«' 
or,  changing  signs  in  the  terms  so  as  to  have  the  same  denominator  as 
the  value  of  X,  we  have 

ac'  —  a'c 
ab'  —  a'b 


EQUATIONS   WITH  THREE  UNKNOWN  QUANTITIES.    191 

Solve  the  following  literal  equations  by  either  method  of 
elimination : 

2.  x  -f-  y  =  a  +  b,  bx  +  ay  =  2t(b.      Ans.  x  =  a,  y  =  b. 

3.  (a  +  c)x  —  by  =  be,  x  +  y  =  a  +  b.       x  =  b,  y  =  a, 

a  7  /         j \  ac  be 

4.  a  +  y  =  c,aa;  —  by  =  c(a  —  b).     x=  -—  ,y=_ __. 

a  +  6  a  +  w 

-     x    ,    y       .    x    ,    y        ,  «£>  (t6 

a       b  b       a  a  +  b  a  +  b 

a      b  b      a  a2  +  b~    J        a2  +  62 

7.    The  sum  of  two  numbers  is  a  and  their  difference  is  6  : 

find  the  numbers.  Ans.  Greater  -  +  -  ;   less . 

When  the  known  quantities  in  a  problem  are  represented 
by  letters,  the  answer  furnishes  a  general  result  or  Formula 
(Art.  41);  and  a  formula  expressed  in  ordinary  language, 
furnishes  a  Rule.  Thus,  in  the  present  example,  we  have 
the  following 

Rule.  The  sum  and  difference  of  two  numbers  being  given, 
to  find  the  numbers :  The  greater  number  is  equal  to  half  the 
sum  plus  half  the  difference ;  the  less  number  is  equal  to  half 
the  sum  minus  half  the  difference. 

100.  Simultaneous  Equations  with  Three  or  More 
Unknown  Quantities.  —  In  order  to  solve  simultaneous 
equations  which  contain  two  unknown  quantities,  we  have 
seen  that  we  must  have  two  equations  (Art.  93).  Similarly 
we  find  that  in  order  to  solve  simultaneous  equations  which 
contain  three  unknown  quantities,  we  must  have  three  equa- 
tions. And  generally,  when  the  values  of  several  unknown 
quantities  are  to  be  found,  it  is  necessary  to  have  as  many 
simultaneous  equations  as  there  are  unknown  quantities. 

Simultaneous  simple  equations  involving  three  or  more 
unknown  quantities,  may  be  solved  by  either  of  the  three 
methods  of  elimination  explained  in  the  preceding  articles ; 


192    EQUATIONS   WITH  Til  RICE  UNKNOWN  QUANTITIES. 

but  the  most  convenient  method  of  elimination  is  generally 
that  by  addition  or  subtraction.     The   unknown  quantities 
are  to  be  eliminated  one  at  a  time  by  the  following 
Rule. 

If  there  be  three  simple  equations  containing  three  unknown 
quantities,  eliminate  one  of  the  unknoum  quantities  from  any 
two  of  the  equations,  by  the  methods  already  explained  (Arts. 
95,  96,  97)  ;  then  eliminate  the  same  unknown  quantity  from 
the  third  given  equation  and  either  of  the  former  two;  two 
equations  involving  two  unknoum  quantities  are  thus  obtained, 
and  the  values  of  these  unknown  quantities  may  be  found  by 
the  rules  given  in  the  preceding  Articles.  The  remaining 
unknown  quantity  may  be  found  by  substituting  these  values 
in  any  one  of  the  given  equations. 

If  four  equations  are  given  involving  four  unknown  quan- 
tities, one  of  the  unknown  quantities  must  be  eliminated 
from  three  pairs  of  the  equations.  Three  equations  involv- 
ing three  unknown  quantities  will  thus  be  obtained,  which 
may  be  solved  according  to  the  rule.  If  five  or  more  equa- 
tions are  given,  they  may  be  solved  in  a  similar  manner. 

Note  1.  —  Either  of  the  unknown  quantities  may  be  selected,  as 
the  one  to  be  first  eliminated;  but  it  is  best  to  begin  with  the  quantity 
which  has  the  simplest  coefficients;  and  when  an  unknown  quan- 
tity is  not  contained  in  all  the  given  equations,  it  is  generally  best 
to  eliminate  that  quantity  first. 

EXAMPLES. 

1.    Solve  6a;  +  2y  -  5«  =  13, (1) 

3a;  +  By  -  2«  =  13 (2) 

7x  +  By  -  Zz  =  26 (3) 

Choose  y  as  the  first  quantity  to  be  eliminated. 
Multiply  (1)  by  3,  and  (2)  by  2, 

18a?  +  Gy  -  15z  =  39, 

<;.,:  +  c,//  _     lv  =  26. 

subtracting,  12*  —  11*  =  13 (O 


EXAMPLES.  193 

Multiply  (1)  by  5,  and  (3)  by  2, 

30a;  +  lOy  -  252  =  65, 
Ux  +  10//  -     62  =  52. 

subtracting,  16a;  —  19z  =  13 (5) 

We  have  now  to  find  the  values  of  x  and  z  from  (4)  and  (5) . 
Multiply  (4)  by  4  and  (5)  by  3  (Art.  95,  Kern.), 

48a;  -  442  =  52, 

48a;  —  572  =  39. 


subtracting,  132  =13.         .-.     2  =  1. 

Substitute  this  value  of  z  in  (4)  ;  thus 

12a:  -  11  =  13.         .-.     x  =  2. 
Substitute  these  values  of  x  and  z  in  (1)  ;  thus 
12  +  2y  —  5  =  13. 
.-.     y=     3,  1 
and  2  =     1,  [ 


Note  2.  —  Although  the  method  of  elimination  given  by  the  rule 
is  generally  the  best,  yet  in  particular  examples  solutions  may  be 
obtained  more  easily  and  elegantly  by  other  means,  which  the  student 
in ust  learn  by  experience.  After  a  little  practice  he  will  find  that  the 
solution  may  often  be  considerably  shortened  by  a  suitable  combination 
of  the  given  equations.     Thus,  Ex.  1  may  be  solved  as  follows: 

Add  (1)  and  (2)  and  subtract  (3), 
2x  -  42  =  0, 

or  x  =  22 (6) 

Substitute  this  value  of  x  in  (1)  and  (2),  and  we  get 
2)j  +  72  =  13, 
3?/  +  Az  =  13. 
Subtracting,  y  —  oz  =     0. 

.-.     y  =  3z (7) 

Substitute  these  values  of  x  and  y  in  (1)  ;  thus 
122  +  62  -  52  =  13. 
.-.     2  =     1; 
therefore  from  (6)  and  (7),  a;  =     2, 

y=    3. 


194  EXAMPLES. 


i-f. (2) 

a;        3?/ 

!-i+*->ft (3) 

a;       5//        z 

Clearing  of  fractional  coefficients,  we  obtain 

from  (1)  §   +^   -    *  =  3, (1) 

aj        ?/  2 

from  (2)  -   -  i  =0, (5) 

a;        ?/ 

from  (3)  —  -  -  +  —  =  32 (6) 

x         if         z 

Choose  z  as  the  first  quantity  to  be  eliminated  (Note  1). 
Multiply  (4)  by  15  and  add  the  result  to  (6)  ;  thus 

M  +  i?  =  77. 

x  y 

Divide  by  7,  —  +   -  =  11 CO 

x         y 

Multiply  (5)  by  6,  —  -    -  =     0. 

x  y 

...  »_ii. 

x 

x  =     3,  ] 
from  (5)  n  =     1,  J 

from  (4)  z  =     2.  j    ■ 

3.    .Solve  5a;  -  3y  -    z  =     6, (1) 

13a;  -  ly  +  3z  =  14, (2) 

7a;  -  4?/  =8 (3) 

Multiply  (1)  by  3  and  add  the  result  to  (2), 

28a;  -  IGy  =  32. 
Divide  by  4,  7a;  —     Ay  =     8. 

Thus  we  see  that  the  combination  of  equations  (1)  and 
(2)  leads  to  an  equation  which  is  identical  with  (3)  ;  and  so 
to  find  x  and  //,  we  have  but  a  single  equation,  7x  —  -1//  =  8, 


EXAMPLES.  195 

with  two  unknown  quantities,  which  is  not  sufficient  to 
determine  the  deiiuite  value  of  either  (Art.  93).  The 
anomaly  here  arises  from  the  fact  that  one  of  these  three 
equations  is  deducible  from  the  others  ;  in  other  words,  that 
the  three  equations  are  not  independent  (Art.  93) . 

Note  3.  —  Sometimes  it  is  convenient  to  use  the  following  rule : 
Express  the  values  of  two  of  the  unknown  quantities  from  two  of 
the  equations  in  terms  of  the  third  unknown  quantity,  and  substitute 
these  values  in  the  third  equation.     From  this,  the  third  unknown 
quantity  can  be  found,  and  then  the  other  two;  thus 

4.  Solve  3;c  4-  Ay  -  I62  =  0, (1) 

5x  -  8y  +  IO2  =  0, (2) 

2x  +  6y  +     72  =  52 (3) 

Multiply  (1)  by  2  and  add  to  (2)  ;  thus 

11a  -  22z  =  0.  .-.     x  =  2z. 

Multiply  (1)  by  5,  and  (2)  by  3,  and  subtract;  thus 

My  -  110a  =  0.  .-.     y  =  — . 

Substitute  these  values  of  x  and  y  in  (3)  ;  thus 

Az  +  15a  +  Iz  =  52. 

.-.      2  =  2,1 
and  x  =  4,  | 

y  =  5.  J 

Note  4.  —  The  rule  in  Note  3  is  especially  convenient  when  all  of 
the  unknown  quantities  occur  in  only  one  equation;  thus 

5.  Solve  x  +  y  +  z  =  a  +  b  +  c,      .     .     .     .   (1) 

x  —  y  =  b  —  «, (2) 

x  —  z  =  c  —  a (3) 

From  (2)  we  have  y  =  x  +  a  —  b (4) 

From  (3)  we  have  2  =  x  +  a  —  c (5) 


19G  PROBLEMS  LEADING  TO  SIMULTANEOUS   EQUATIONS. 


Substitute  these  values  of  y  and  z  in  (1), 

x  +  x  +  a  —  b  +  x  +  a  —  c  =  a  +  b  +  c. 

...      33  =   _a  +  26  +  2c. 

...       3.=  |(a  +  6  +  c)  _ 

from  (4) 

from  (5) 

Solve  the  following  equations 

G.  7»  +  3y  -  2z  =  1G, 
2x  +  5y  +  3z  =  39, 
5a;  —    y  +  5z  =  31. 

7.  2xc  +  32/  +  4z  =  16, 
3a  +  2y  —  5z  =    8, 

5a  —  Gy  +  3z  =     6. 


."' 


x  +  2y  +  2z  = 

11, 

2x  +     y  +     g  = 

7, 

Bx  +  4?/  +     a  = 

14. 

a  +  3?/  +  \z  = 

14, 

x  +  2//  +     z  = 

7, 

2x  +    y  +  2z  = 

2. 

1          2  _    3  _ 
x         y        z 

1, 

x         y         z 

24, 

7  _    8         9  _ 

14. 

|(a  +6  +  c 

)   - 

-  &, 

|(a  +  & .+  < 

)   " 

-  c. 

J 

r. 

. 

2, 

yl»s. 

= 

4, 

>« 

= 

5. 

.r 

= 

3, 

2/ 

= 

2. 

2 

= 

1. 

'  a; 

= 

1. 

2/ 

= 

2, 

z 

= 

3. 

a; 

=  - 

-2, 

2/ 

= 

4, 

z 

= 

1. 

\  y  = 


101.  Problems  Leading  to  Simultaneous  Equa- 
tions.—  AVe  shall  now  give  some  examples  of  problems 
which  lead  to  simultaneous  equations  of  the  first  degree  with 
two  or  more  unknown  quantities.  Many  of  the  problems 
given  in  Chapter  IX.  really  contain  ttvo  or  more  unknown 
quantities,  but  the  given  relations  are  there  of  so  simple  a 
nature  that  it  is  easy  to  express  all  of  the  unknown  quanta- 


EXAMPLES.  197 

ties  in  terms  of  one  unknown  quantity,  and  thus  to  require 
but  a  single  equation.  In  the  problems  of  the  present 
chapter  the  relations  between  the  unknown  quantities  are 
not  so  simple,  and  the  solution  will  give  rise  to  simultaneous 
equations  ;  and  in  all  cases  the  conditions  of  the  problem 
must  be  sufficient  to  furnish  as  many  independent  equations 
as  there  are  unknown  quantities  to  be  determined  (Art.  100) . 

EXAMPLES. 

1.    Find  two  numbers  such  that  the  greater  exceeds  twice 
the  less  by  3,  and  that  twice  the  greater  exceeds  the  less  by 


27. 

Let  x  = 

the  greater  number, 

and      y  = 

the  less  number. 

Then  from  the  conditions, 

x  — 

2y  = 

3, 

and 

2x  - 

y  = 

27. 

Solving  these  equations,  we  have  x  =  17,  y  =  7. 

If  the  numerator  of  a  fraction  be  increased  by  2  and 
the  denominator  by  1,  it  becomes  equal  to  f ;  and  if  the 
numerator  and  denominator  are  each  diminished  by  1,  it 
becomes  equal  to  £ :  find  the  fraction. 

Let  x  =  the  numerator, 
and      y  =  the  denominator. 

Then  from  the  conditions, 

X_±_2_    5 
ii     —    8' 

y  +  i 

and  ^^i  =  i. 

y  -  1 

Solving,  we  have  x  =  8,  y  =  15. 
Hence  the  fraction  is  ^. 

(^3y  -A-  man  and  a  boy  can  do  in  15  days  a  piece  of  work 
which  would  be  done  in  2  days  by  7  men  and  9  boys  :  how 
long  would  it  take  one  man  to  do  it? 


198  EXAMPLES. 

Let    x  =  the  number  of  days  in  which  one  man  would  do 
the  whole, 
and  let  y  =  the  number  of  days  in  which  one  boy  would  do 
the  whole. 

Then     -  =  the  part  that  one  man  can  do  in  one  day, 
x 

and        -  =  the  part  that  one  boy  can  do  in  one  day. 

y 

Then  from  the  conditions   of  the  question,  a  man  and  a 
boy  together  do  TV^n  °f  the  wol'k  in  one  day  ;  hence  we  have 

i  +  i  =  A 0) 

x       y 
Also,  since  7  men  and  9  boys  do  half  the  work  in  a  day, 
we  have  „       0 

-  +  -  =  \ (2) 

x       y 

Multiplying  (1)  by  9,  and  subtracting  (2)  from  it,  we  have 
2 
x 

Thus  one  man  would  do  the  work  in  20  days. 
4.  A  railway  train  after  traveling  an  hour  is  detained  24 
minutes,  after  which  it  proceeds  at  six-fifths  of  its  former 
rate,  and  arrives  15  minutes  late.  If  the  detention  had 
taken  place  5  miles  further  on,  the  train  would  have  arrived 
2  minutes  later  than  it  did.  Find  the  original  rate  of  the 
train,  and  the  distance  traveled. 

Let  x  =  the  original  rate  of  the  train  in  miles  per 

hour ; 
and  y  =  the  number  of  miles  in  the  whole  distance 

traveled. 
Then      y  —  x  =  the  number  of  miles  to  be  traveled  after 
the  detention. 
^  ~  '    =  the  number  of   hours  in  traveling  y  —  x 
miles  at  the  original  rate, 
and  — -~ — -  =  the  number  of  hours  in  traveling  y  —  x 
miles  at  the  increased  rate. 


&.         .-.     s=20. 


EXAMPLES.  199 

Since  the  train  is  detained  24  minutes,  and  yet  arrives  only 
15  minutes  late,  it  follows  that  the  remainder  of  the  journey 
is  performed  in  nine  miuutes  less  than  it  would  have  been  if 
the  rate  had  not  been  increased  ;  hence  we  have 

>        x  6x         ~^° (1) 

If  the  detention  had  taken  place  5  miles  further  on,  there 
would  have  been  y  —  x  —  5  miles  left  to  be  traveled  after 
the  detention  :  hence  we  have 


Subtract  (2)  from  (1), 


5  (.7  -  x  —  5) 
6x 
5  _  25 
x       Co 

6x 


(2) 


.*.     x  =  25, 

an(*  y  =  47£. 

^v 'There  is  a  number  consisting  of  three  digits ;  the 
middle  digit  is  zero,  and  the  sum  of  the  other  digits  is  11  ; 
if  the  digits  be  reversed,  the  number  so  formed  exceeds  the 
original  number  by  495  :  find  the  number. 

Let  x  =  the  digit  in  the  unit's  place, 
and      y  =  the  digit  in  the  hundred's  place. 

Then,  since  the  digit  in  the  ten's  place  is  0,  the  number 
will  be  represented  by  100?/  +  x  (Art.  92,  Ex.  10)  ;  hence 
from  the  conditions,  we  have 

x  +  y  =     11, 

and  100a  +  y  —  (100//  +  x)  =  495. 

Solving,  we  get  x  =  8,  y  =  3  ;  hence  the  number  is  308. 
^6j  A  sum  of  money  was  divided  equally  among  a  certain 
number  of  persons  ;  had  there  been  three  more,  each  would 
have  received  SI  less,  and  had  there  been  two  less,  each 
would  have  received  $1  more  than  he  did  :  find  the  number 
of  persons,  and  what  each  received. 


200  EXAMPLES. 

Let     x     =  the  number  of  persons, 
and  y  =  the  number  of  dollars  which  each  received. 

Then  xy  =  the  number  of  dollars  to  be  divided,  and  from 
the  conditions,  wc  have 

(x  +  3)0/  -  1)  =  xy, 

and  {x-  2){y  +  1)  =  xy. 

Solving,  we  get  x  =  12  and  y  =  5. 

7.  A  train  traveled  a  certain  distance  at  a  uniform  rate  ; 
had  the  speed  been  6  miles  an  hour  more,  the  journey  would 
have  occupied  4  hours  less  ;  and  had  the  speed  been  6  miles 
an  hour  less,  the  jouruey  would  have  occupied  G  hours  more  : 
find  the  distance. 

Let     x    =  the  rate  of  the  train  in  miles  per  hour, 
and  y  —  the  time  of  running  the  journey  in  hours. 

Then  xy  =  the  distance  traversed,  and  from  the  conditions, 
we  have  (x  +  6)  (y  —  4)  =  xy, 

and  (x  -  6)  (y  +  6)  =  xy. 

Solving,  we  get  x  =  30,  and  y  =  2-4.     Hence  the  distance 
is  720  miles. 

8.  A,  B,  and  C  can  together  do  a  piece  of  work  iu  30 
days ;  A  and  B  can  together  do  it  in  32  days  ;  and  B  and  C 
can  together  do  it  in  120  days  :  find  the  time  in  which  each 
alone  could  do  the  work. 

Let  x  =  the  number  of  days  in  which  A  could  do  it, 
y  =  the  number  of  days  in  which  B  could  do  it, 
and      z  =  the  number  of  days  in  which  C  could  do  it. 
Then  we  have  from  the  conditions, 

1  +1  4-  1  -    i 
so      y      z 

1,1 

— I —  =  SZi 

x       y 

l.l-i 

y  +  -z-^~- 

Solving,  we  get  x  =  40,  y  =  160,  z  =  480. 


EXAMPLES.  201 

Find    the    fraction   which    is    equal    to   ^   when   its 

ferator  is  increased  by  unity,  and  is  equal  to  \  when 

its  denominator  is  increased  by  unity.  Ans.  §. 

JlO>)  A  certain  number  of  two  digits  is  equal  to  five  times 

the  sum  of  its  digits;    and  if  nine  be  added  to  the  number 

the  digits  are  reversed  :  find  the  number.  Ans.  45. 

11.  If  15  lbs.  of  tea  and  17  lbs.  of  coffee  together  cost 
$7.86,  and  25  lbs.  of  tea  and  13  lbs.  of  coffee  together 
cost  $10.34,  find  the  price  per  pound  of  each. 

Ans.  The  tea  cost  32  cents,  and  the  coffee  cost  18  cents, 
alb. 

12.  If  A's  money  were  increased  by  $36  he  would  have 
three  times  as  much  as  B  ;  and  if  B's  money  were  diminished 
by  $5  he  would  -have  half  as  much  as  A  :  find  the  sum 
possessed  by  each.  Ans.  A  has  $42,  B  has  $26. 

13.  Find  two  numbers  such  that  half  the  first  with  a  third 
of  the  second  may  make  32,  and  that  a  fourth  of  the  first 
with  a  fifth  of  the  second  may  make  18.  Ans.  24,  60. 

14.  A  farmer  parting  with  his  stock,  sells  to  one  9  horses 
and  7  cows  for  $1200  ;  and  to  another,  at  the  same  prices, 
6  horses  and  13  cows  for  the  same  sum  :  what  was  the  price 
of  each?  Ans.  $96,  $48. 

15.  Having  $45  to  give  away  among  a  certain  number  of 
persons,  I  find  that  for  a  distribution  of  $3  to  each  man  and 
$1  to  each  woman,  I  shall  have  $1  too  little  ;  but  that,  by 
giving  $2.50  to  each  man  and  $1.50  to  each  woman,  I  may 
distribute  the  sum  exactly  :  how  many  were  there  of  men 
and  women?  Ans.  12,  10. 

16.  Find  three  numbers,  A,  B,  C,  such  that  A  with  half 
of  B,  B  with  a  third  of  C,  and  C  with  a  fourth  of  A,  may 
each  be  1000.  Ans.  640,  720,  840. 

17.  A  person  spent  $1.82  in  buying  oranges  at  the  rate  of 
3  for  two  cents,  and  apples  at  5  cents  a  dozen  ;  if  he  had 
bought  five  times  as  many  oranges  and  a  quarter  of  the 
number  of  apples  he  would  have  spent  $5.30  :  how  many  of 
each  did  he  buy?  Ans.  153,  192. 


202  EXAMPLES. 

EXAMPLES. 

Solve  the  following  equations  : 

1.  bx  —  ly  =  0,  Ix  +  by  =  74.      Ar.s.  x  =  7,  y  =  5. 

2.  5as  =  7?/  —  21,  21.r  —  9?/  =  75.  x  =  7,  y  =  8. 

3.  6?/  -  5aj  =  18,  12a:  -  9y  =  0.  a;  =  G,  ?/  =  8. 

4.  7.x  +  4y  =  1,  9.f  +  Ay  =  3.  a?  =  1,  y  =  -1|. 

5.  x  -  Uy  —  1,  Illy  —  9x  =  99.        x  =  100,  ?/  =  9. 
G.    8x  —  21y  =  5,  6x  +  Uy  =  —  26.       £c=—  2,2/=  —  1. 

7.  39a;  -  8#  =  99,  52a;  -  I5y  =  80.       x  =  5,  y  =  12. 

8.  3a;  =  7?/,  12?/  =  5a?  —  1.  x  =  —7,  y  =  —3. 

9.  93a;  +  15?/  =  123,  15a:  +  93?/  =  201.       x  =  1,  ?/  =  2. 

10.  ?  +  £  =  i  2  _  ?y  =  3.  x  =  A  y  =  _3> 

2       3  4         3  J 

11.  £_±J  +  x  =  15,  EJZJ  +  y  =  6.        «  =  10,  3,  =  5. 

3  5 

12.  ^  +  ^  =  34,^  +  ^  =  ^+12.       ^=12^  =  12. 

6         3  84         8 

10     l-3a;  ,  3v  — 1      0  3a;  +  w  Q  „ 

13.  _      +    *         =2, —-p£  +  y  =  9.        a?=o,y  =  7. 

7  5  11 

14.  |-i(y-2)  -i(*-3)  =0, .,_  t  (<y_  l)  -  i  (x-2)  =0. 

^l//,s'.  x  =  3f ,  ?/  =  C>f . 

ir     cc-2       ?/  +  2       n  2a;— 5       11—2?/       ,.  -  , 

15. Z-I—  =  0, •    =  0.      #=5,  y  =  2. 

3  4  5  7 

16.  -  +  -^  =  3a:-7?/-37,  3a:-7?/=37.        a;=3,?/  =  -4. 
o       4 

1 7.  (a;+l)  0/+5)  =  (aj+5)  (//+ 1 ),  «y+a;+y=  (a?+2)  (*/+2). 

Ans.  x  =  —2,y  =  —2. 

18.  xy  -  (y  -  l)(x  -  1)  =  6(.y  -  \),x-y=  1. 

.Aws.  .i-  =  2h  y  =  l  J. 

19.  l±*  =  1+J  =  n  +•»  +  *  x  =  8,y=  L6 

3  5  7  '* 


EXAMPLES. 


203 


20. 

21. 
22. 

23. 

24. 

26. 

26. 

27. 
28. 

29. 
30. 
31. 

32. 

33. 


,3x  +  .125?/  =  x  —  6,  3oj  —  .5*/  =  28  -  .25*/. 

^4»is.  a;  =  10,  y  =  8. 
.08<c-.21y  =  .33,  .12a;  +.7?/  =  3.54.        x  =  12,  y  =  3. 


l!_:U2,i8  +  8^io. 

a;       y  x        y 

?  +  «_7,»-*-ll. 

a;  +  5  =  *   3iC  _  ?  =  26. 

y  y 

2x  -  £  =  3,  8a;  +  —  =  - 

2/  2/ 


«  =  24,  Z/  =  2f. 

*  =  «.*-  -i». 

a;  =  3,  ?/  =  6. 


•6. 


a;  —  5  6  4 


*  +  ^  =  2,  6a;  -  aw  =  0. 
a       b 


Ans.  x  =  4,  y  =  12. 

x  =  a,  y  =  b. 


a(x  +  y)  +  b(x  -  a)  =  1,  a(x  -y)  +  b(x  +  y)  =  1 . 

1 


Ans.  x 


,y  =  0. 


a  +  b 
x  +  y  =  a  +  b,  ax  —  by  +  cr  —  b2  =  0. 

Ans.  x  =  2b  —  a,  y  =  2a  —  b. 
{a+b)x  +  (a-b)y  =  2ac,  (&+c)a  +  (6— c)?/  =  26c. 

Ans.  x  =  y  =  c. 
a;  +  2?/  —  32  =    6,  f  x  =      8|, 

2a;  +  4y  -  72  =    9,  //  =      3£, 

3a;  —    y  —  52  =    8.  [  2  =      3. 

2a;  -    y  +    2  =    4,  \x=  -1, 

5a;  +     ?/  +  3z  =     5,  {  y  =  -2, 


2a;  -  32/  +  iz  =  20. 

aj  4.  Ay  +  3z  =  17, 

3a,  +  Sy  +    2  =  16, 

2x  +  2y  +    2  =  11. 


[  2    =a  4. 

'  a;  =  2, 

2/  =  3, 

{z  =  1. 


204 


EXAMPLES. 


34. 

35. 

3G. 

37. 
38. 
39. 


2a:  +  By  +  \z  —  20, 
3a;  +  4//  +  hz  =  26, 
3a:  +  by  +  62  =  81. 


-4?is. 


J- 


V 


6,y-; 


»,  2- 


40. 


41, 


42. 


43. 


II. 


aj       y       2 


lu. 


c  _  1    2a 

■  -  —  1 , 
2  a; 

.4ns.  a;  = 


:8,  y=10,  2=14. 

-  *  -  2  =  b. 

7/  0 

a,  y  =  b,  z  =  c. 


1  +  1=1,1  +  1=2,1  +  1  = 

?+!-?,  »-*=*,  !+!=*. 

x    y     z   z     y         x     z 


4,z 


:;,/ 


5a:    ,    4z 

T     "3 


6z       aj   ,    „ 

"5"  "  2  +  f ' 


2/  + 


1  .'/  =     3, 


3a;  +  1 z_ 

7  14 


22  +  *. 
21       3 


7a;  —  3//  =  1,  llz  —  7«  =  1, 
Az  —  ly  =  1,  19a;  —  3w  =  1. 

3m  -  2y  =    2,5a;  —  Iz  —  11. 
2a;  +  Sy=  39, 4v  +  3«  =11, 

2a;— 3v-f  22  =  13,4^+22=14. 
4%-2a;=30,  5v+3m=32. 

7a;— 22+3w=17,4y— 22+v=ll, 
5?/ — 3a;— 2«= 8,  4y — 3 u  +  2 v  =  9 , 
32+8m=33. 

3a;  -  1//  +  32  +  Sv  —  6it  =.  11, 
Sx  —  by  +  22  —  4m  =  11, 
10//  -  ?>z  +  3*t  —  2v  =  2, 
52  -f  4m  +  2«  -  2a;  =  3, 
Ga  —  00  +  l.o  —  'ly  =  6. 


(  •>•=   4,  M= 

(  2  =  16,  M= 
(M  = 


4,  a;: 

5,2: 


x=  3,y= 

?<=   i),z  = 


1    2: 


2,  //  = 

3,  w= 


12, 

7. 

1, 

5. 

4, 
3, 


[v=   1. 


=  — 1, 


V 


EXAMPLES.  205 

45.  What  fraction  is  that,  to  the  numerator  of  which  if  7 
be  added,  its  value  is  f  ;  but  if  7  be  taken  from  the  denom- 
inator its  value  is  f  ?  Ans.  ^. 

46.  A  rectangular  bowling-green  having  been  measured, 
it  was  observed  that,  if  it  were  5  feet  broader  and  4  feet 
longer,  it  would  contain  116  feet  more;  but  if  it  were  4 
feet  broader  and  5  feet  longer,  it  would  contain  113  feet 
more  :  find  its  area.  Ans.  108  sq.  ft. 

47.  A  party  was  composed  of  a  certain  number  of  men 
and  women,  and,  when  four  of  the  women  were  gone,  it  was 
observed  that  there  were  left  just  half  as  many  men  again  as 
women  ;  they  came  back,  however,  with  their  husbands,  and 
now  there  were  only  a  third  as  many  men  again  as  women  : 
what  were  the  original  numbers  of  each?  Ans.  12,  12. 

48i  The  sum  of  the  two  digits  of  a  certain  number  is  6 
times  their  difference,  and  the  number  itself  exceeds  6  times 
their  sum  by  3  :  find  the  number.  Ans.  75. 

49.  Divide  the  numbers  80  and  90  each  into  two  parts,  so 
that  the  sum  of  one  out  of  each  pair  may  be  100,  and  the 
difference  of  the  others  30. 

Ans.  30,  50,  and  70,  20  ;  or  60,  20,  and  40,  50. 

50.  Four  times  B's  age  exceeds  A's  age  by  20  years,  and 
one-third  of  A's  age  is  less  than  B's  age  by  2  years  :  find 
their  ages.  Ans.  A  36  years,  B  14  years. 

^-  §if.  In  8  hours  A  walks  12  miles  more  than  B  does  in  7 
brours  ;  and  in  13  hours  B  walks  7  miles  more  than  A  does 
in  9  hours :  how  many  miles  does  each  walk  per  hour? 

.-^  Ans.  A  5  miles,  B  4  miles. 

[52/  The  sum  and  the  difference  of  a  number  of  two  digits 
ana  of  the  number  formed  by  reversing  the  digits  are  110 
and  54  respectively:  find  the  numbers.  Ans.  28,  82. 

53.  In  a  bag  containing  black  and  white  balls,  half  the 
number  of  white  is  equal  to  a  third  of  the  number  of  black  ; 
and  twice  the  whole  number  of  balls  exceeds  three  times  the 
number  of  black  balls  by  four :  how  many  balls  did  the  bag 
contain?  Ans.  8  white,  12  black. 


206  EXAMPLES. 

54.    Twenty-eight  tons  of  goods  arc  to  be  carried  in  carts 
and  wagons,  and  it  is  found  that  this  will  require  15  carts  and 
12  wagons,  or  else  24  carts  and  8  wagons:  how  much  can 
..each  cart  and  each  wagon  carry?  Ans.  §  tons,  f  tons. 

•-*•  55.  The  first  edition  of  a  book  had  600  pages,  and  was 
divided  into  two  parts  ;  in  the  second  edition  one  quarter  of 
the  second  part  was  omitted  and  30  pages  added  to  the  first ; 
the  change  made  the  two  parts  of  the  same  length  :  what 
were  they  in  the  first  edition?  Ans.  240,  360. 

56.  If  A  were  to  receive  $10  from  B  he  would  then  have 
twice  as  much  as  B  would  have  left ;  but  if  B  were  to  receive 
$10  from  A,  B  would  have  three  times  as  much  as  A  would 
have  left:  how  much  has  each?  Ans.  $22,  $26. 

57.  A  farmer  sold  30  bushels  of  wheat  and  50  bushels  of 
barley  for  $75  ;  he  also  sold  at  the  same  prices  50  bushels 
of  wheat  and  30  bushels  of  barley  for  $77:  what  was  the 
price  of  the  wheat  per  bushel?  Ans.  $1. 

58.  A  certain  fishing  rod  consists  of  two  parts  ;  the  length 
of  the  upper  part  is  to  the  length  of  the  lower  as  5  to  7  ; 
and  9  times  the  upper  part  together  with  13  times  the  lower 
part  exceeds  11  times  the  whole  rod  by  36  inches  :  find  the 
lengths  of  the  two  parts.  Ans.  45,  63. 

59.  A  certain  company  in  a  tavern  found,  when  they  came 
to  pay  their  bill,  that  if  there  had  been  3  more  persons  to 
pay  the  same  bill,  they  would  have  paid  $1  each  less  than 
they  did ;  and  if  there  had  been  2  fewer  persons  they  would 
have  paid  $1  each  more  than  they  did :  find  the  number 
of  persons,  and  the  number  of  dollars  each  paid. 

Ans.  12,  5. 

60.  There  is  a  rectangular  floor,  such  that  if  it  had  been 
2  feet  broader,  and  3  feet  longer,  it  would  have  been  61 
square  feet  larger;  but  if  it  had  beeo  3  feet  broader,  and  2 
feet  longer,  it  would  have  been  68  square  feet  larger :  find 
the  length  and  breadth  of  the  floor.  Ans.  14  ft.,  10  ft. 

Let  x  =  the  length,  and  y  —  the  breadth,  of  the  floor  in  feet;  then 
.nj  —  the  surface  of  the  floor  In  square  feet. 


EXAMPLES.  207 

61.  "When  a  certain  number  of  two  digits  is  doubled,  and 
increased  by  3G,  the  result  is  the  same  as  if  the  number  had 
been  reversed,  and  doubled,  and  then  diminished  by  36  ; 
also  the  number  itself  exceeds  4  times  the  sum  of  its  digits 
by  3  :  find  the  number.  Ans.  59. 

62.  Two  passengers  have  together  560  lbs.  of  luggage, 
and  are  charged  for  the  excess  above  the  weight  allowed  62 
cents  and  $1.18  respectively  ;  if  the  luggage  had  all  belonged 
to  one  of  them  he  would  have  been  charged  $2.30 :  how 
much  luggage  is  each  passenger  allowed  without  charge  ? 

Ans.  100  lbs. 

63.  A  farmer  has  28  bushels  of  barley  at  56  cents  a 
bushel ;  with  these  he  wishes  to  mix  rye  at  72  cents  a  bushel, 
and  wheat  at  96  cents  a  bushel,  so  that  the  mixture  may 
consist  of  100  bushels,  and  be  worth  80  cents  a  bushel :  how 
many  bushels  of  rye  and  wheat  must  he  take?     Ans.  20,  52. 

64.  A  and  B  ran  a  race  which  lasted  5  minutes  ;  B  had  a 
MJtart  of  20  yards ;  but  A  ran  3  yards  while  B  was  running  2, 

and  won  by  30  yards :  find  the  length  of  the  course  and  the 
rate  of  each  per  minute. 

Ans.  150  yards,  30  yards,  20  yards. 

v;/o5.   A  and  B  can  together  do  a  certain  work  in  30  days ; 

at  the  end  of  18  days  however  B  is  called  off  and  A  finishes 

it  alone  in  20  days  more :  find  the  time  in  which  each  could 

do  the  work  alone.  Ans.  50,  75. 

j\  G6.   A,  B,  and  C  can  together  drink  a  cask  of  beer  in  15 

days ;  A  and  B  together  drink  four- thirds  of  what  C  does  ; 

and  C  drinks  twice  as  much  as  A :  find  the  time  in  which 

eachalone  could  drink  the  cask  of  beer.        Ans.  70,  42,  35. 

,Y.//^7.    A  and  B  run  a  mile  ;  at  the  first  heat  A  gives  B  a 

'start  of  20  yards,  and  beats   him   by  30   seconds;    at   the 

second  heat  A  gives  B  a  start  of  32  seconds,  and  beats  him 

by  9^  yards  :  find  the  rate  per  hour  at  which  A  runs. 

Ans.  12  miles. 

68?    A  and  B  are  two  towns  situated  24  miles  apart,  on 

the  same  bank  of  a  river.     A  man  goes  from  A  to  B  in  7 


208  EXAMPLES. 

hours,  1))T  vowing  the  first  half  of  the  distance,  and  walking 
the  second  half.  In  returning  he  walks  the  first  half  at 
three-fourths  of  his  former  rate,  but  the  stream  being  with 
him  he  rows  at  double  his  rate  in  going  ;  and  he  accomplishes 
the  whole  distance  in  G  hours.  Find  his  rates  of  walking  and 
rowing  up  stream.  Ans.  4  miles  walking,  3  miles  rowing. 
/69.  A  railway  train  after  traveling  an  hour  is  detained  15 
minutes,  after  which  it  proceeds  at  three-fourths  of  its  former 
rate,  and  arrives  24  minutes  late.  If  the  detention  had 
taken  place  5  miles  further  on,  the  train  would  have  arrived 
3  minutes  sooner  than  it  did.  Find  the  original  rate  of  the 
train  and  the  distance  traveled. 

Ans.  33^  miles  per  hour,  48^  distance. 
70;  The  time  which  an  express  train  takes  to  travel  a 
journey  of  120  miles  is  to  that  taken  by  an  ordinary  train  as 
9  to  14.  The  ordinary  train  loses  as  much  time  in  stopping 
as  it  would  take  to  travel  20  miles  without  stopping.  The 
express  train  loses  only  half  as  much  time  in  stoppiug  as  the 
ordinary  train,  and  it  also  travels  15  miles  au  hour  faster. 
Find  the  rate  of  each  train.         •  Ans.  45,  30  miles  per  hour. 

71.  Two  trains,  92  feet  long  and  84  feet  long  respectively, 
are  moving  with  uniform  velocities  on  parallel  rails  ;  when 
they  move  in  opposite  directions  they  are  observed  to  pass 
each  other  in  one  second  and  a  half ;  but  when  they  move 
in  the  same  direction  the  faster  train  is  observed  to  pass  the 
other  in  6  seconds :  find  the  rate  at  which  each  train  moves. 

Ans.  30,  50  miles  per  hour. 

72.  A  railroad  runs  from  A  to  C.  A  freight  train  starts 
from  A  at  12  o'clock,  and  a  passenger  train  at  1  o'clock. 
After  going  two-thirds  of  the  distance  the  freight  train 
breaks  down,  and  can  only  travel  at  three-fourths  of  its 
former  rate.  At  40  minutes  past  2  o'clock  a  collision 
occurs,  10  miles  from  C.  The  rate  of  the  passenger  train 
is  double  the  diminished  rate  of  the  freight  train.  Find  the 
distance  from  A  to  C,  and  the  rates  of  the  trains. 

Ans.  GO  miles,  30  and  20  miles  per  hour. 


EXAMPLES.  209 

73.  If  there  were  no  accidents  it  would  take  half  as  long 
to  travel  the  distance  from  A  to  B  by  railroad  as  by  coach ; 
but  3  hours  beiug  allowed  for  accidental  stoppages  by  the 
former,  the  coach  will  travel  the  distance  all  but  15  miles  in 
the  same  time ;  if  the  distance  were  two-thirds  as  great  as  it 
is,  and  the  same  time  allowed  for  railway  stoppages,  the 
coach  would  take  exactly  the  same  time :  required  the  dis- 

,    tance.      .  Ans.  90  miles. 

74.  A  and  B  start  together  from  the  foot  of  a  mountain 
to  go  to  the  summit.  A  would  reach  the  summit  half  an 
hour  before  B,  but  missing  his  way  goes  a  mile  and  back 
again  needlessly,  during  which  he  walks  at  twice  his  former 
pace,  and  reaches  the  top  6  minutes  before  B.  C  starts  20 
minutes  after  A  and  B  and  walking  at  the  rate  of  two  and 
one-seventh  miles  per  hour,  arrives  at  the  summit  10  minutes 
after  B.  Find  the  rates  of  walking  of  A  and  B,  and  the 
distance  from  the  foot  to  the  summit  of  the  mountain. 

Ans.  2±,  2  miles  per  hour ;  distance  5  miles. 
7."i.  A  railway  train  after  traveling  for  one  hour  meets 
with  an  accident  which  delays  it  one  hour,  after  which  it 
proceeds  at  three-fifths  of  its  former  rate,  and  arrives  at  the 
terminus  3  hours  behind  time  ;  had  the  accident  occurred 
50  miles  further  on,  the  train  would  have  arrived  1  hour  20 
minutes  sooner :  find  the  length  of  the  line,  and  the  original 
rate  of  the  train.         Ans.  100  miles,  rate  25  miles  per  hour. 

76.  The  fore-wheel  of  a  carriage  makes  6  revolutions 
more  than  the  hind- wheel  in  going  120  yards  ;  if  the  circum- 
ference of  the  fore-wheel  be  increased  by  one-fourth  of  its 
present  size,  and  the  circumference  of  the  hind-wheel  by 
one-fifth  of  its  present  size,  the  6  revolutions  will  be  changed 
to  4  :  find  the  circumference  of  each  wheel. 

Ans.  4  yards,  5  yards. 

77.  A  and  B  can  perform  a  piece  of  work  together  in  48 
days  ;  A  and  C  in  30  days  ;  and  B  and  C  in  26f  days  :  find 
the  time  in  which  each  could  perform  the  work  alone. 

■Ans.  120,  80,  40  days. 


210  EXAMPLES. 


7.S.  There  is  a  certain  number  of  three  digits  which  is 
equal  to  48  times  the  sum  of  its  digits  ;  and  if  198  be  sub- 
tracted from  the  number  the  digits  will  be  reversed  ;  also  the 
sum  of  the  extreme  digits  is  equal  to  twice  the  middle  digit : 
find  the  number.  Ans.  432. 

79.  A  man  bought  10  horses,  120  oxen,  and  46  cows. 
The  price  of  3  oxen  is  equal  to  that  of  5  cows.  A  horse, 
an  ox,  and  a  cow  together  cost  a  number  of  dollars  greater 
by  300  than  the  whole  number  of  animals  bought ;  and  the 
whole  sum  spent  was  $9366.  Find  the  price  of  a  horse,  an 
ox,  and  a  cow  respectively.  Ans.  $420,  $35,  $21. 

80.  A  farmer  sold  at  a  market  100  head  of  stock  consist- 
ing of  horses,  oxen,  and  sheep,  so  that  the  whole  realized 
$9.40  per  head  ;  while  a  horse,  an  ox,  and  a  sheep  were  sold 
for  $88,  $50,  and  $6  respectively.  Had  he  sold  one-fourth 
the  number  of  oxen,  and  25  more  sheep  than  he  did,  the 
amount  received  would  have  been  still  the  same.  Find 
the  number  of  horses,  oxen,  and  sheep,  respectively,  which 
he  sold.  Ans.  2,  4,  94. 

81.  Five  persons,  A,  B,  C,  D,  E,  play  at  cards;  after  A 
has  won  half  of  B's  money,  B  one-third  of  C's,  C  one-fourth 
of  D's,  D  one-sixth  of  E's,  they  have  each  $30 :  find  how 
much  each  had  to  begin  with. 

Ans.  A  $11,  B  $38,  C  $33,  D  $32,  E  $36. 

82.  A  offers  to  run  three  times  round  a  course  while  B 
runs  twice  round,  but  A  only  gets  150  yards  of  his  third 
round  finished  when  B  wins.  A  then  offers  to  run  four 
times  round  to  B  three  times,  and  now  quickens  his  pace  so 
that  he  runs  4  yards  in  the  time  he  formerly  ran  3  yards. 
B  also  quickens  his  so  that  he  runs  9  yards  in  the  time  he 
formerly  ran  8  yards,  but  in  the  second  round  falls  off  to  his 
original  pace  in  the  first  race,  and  in  the  third  round  goes 
only  9  yards  for  10  he  went  in  the  first  race,  and  accordingly 
this  time  A  wins  by  180  yards.  Determine  the  length  of  the 
course.  Ans.  600  vanls- 


INDETERMINATE  EQUATIONS.  211 


CHAPTER    XL 

INDETERMINATE     PROBLEMS  —  DISCUSSION 

OF     PROBLEMS  — INEQUALITIES. 

102.  Indeterminate  and  Inconsistent  Equations  — • 

Impossible  Problems. — An  Equation  is  said  to  be  Inde- 
terminate when  it  can  be  satisfied  by  an  indefinite  number 
of  sets  of  values  of  the  unknown  quantities  that  enter  it. 
Thus,  an  equation  containing  two  unknown  quantities  may 
be  satisfied  by  as  many  pairs  of  values  as  we  please  (Art. 
93).     Therefore, 

Every  equation  containing  tivo  unknown  quantities  is  inde- 
terminate. 

If  there  are  two  equations  containing  three  unknown 
quantities,  we  may  eliminate  one  of  them,  and  reduce  the 
two  equations  to  a  single  equation  containing  two  unknown 
quantities,  which  we  have  already  seen  is  indeterminate. 

Similarly,  if  there  is  any  number  of  equations  containing 
more  unknown  quantities  than  there  are  equations,  we  may  by 
elimination  reduce  them  to  a  single  equation  containing  two  or 
more  unknown  quantities,  which  we  have  already  seen  is  inde- 
terminate.    Hence  we  have  the  following  general  principle  : 

Any  collection  of  equations  containing  more  unknown 
quantities  than  there  are  equations  can  be  reduced  to  a  single 
indeterminate  equation. 

A  Problem  is  said  to  be  Indeterminate  when  it  imposes 
fewer  conditions  than  there  are  unknown  quantities,  that  is, 
when  it  leads  to  a  less  number  of  independent  equations 
than  there  are  unknown  quantities. 

A  problem  which  involves  only  one  unknown  quantity  may 
be  indeterminate,  the  equation  which  it  leads  to  being  an 
identical  one  (Art.  52).     Thus, 


212  INDETERMINATE   EQUATIONS. 

What  number  is  that  whose  |  increased  by  the  \  is  equal 
to  the  ^7()  increased  by  the  ^  ? 
Let  a;  =  the  number. 

Then  -  +  -  =  *V  +  —  5 

5        G        20         GO' 

clearing  of  fractious,  12a;  +  IOj;  =  21a;  +  x, 

or  22a;  =  22a;, 

which  can  be  satisfied  by  any  value  whatever  of  x. 

It  is  uot  always  the  case  that  the  values  of  two  unknown 
quantities  can  be  found  from  two  independent  equations. 
Thus,  take  the  equations 

x  +  3y  =  4, (1) 

2.x-  +  Gy  =  G (2) 

Multiply  (1)  by  2,  2x  +  6y  =  8, 
and  from  (2),  2a  +  6?/  =  G.         .-.     0  =  2, 

an  impossible  result,  which  shows  that  there  are  no  values 
of  x  and  y  which  will  satisfy  both  (1)  and  (2).  Such 
equations  are  called  Inconsistent  Equations. 

When  a  problem  leads  to  a  greater  number  of  independent 
equations  than  there  are  unknown  quantities,  it  is  Impossible. 
Thus,  suppose  we  have  a  problem  furnishing  the  three  inde- 
pendent equations 

x  +  y  =  12, (l) 

2x  +  y  =  17,    .    .    *    .    .    •  (2) 
SX  +  y=28 (3) 

The  values  of  x  and  y  from  (1)  and  (2)  are  x  =  5,  y  =  7  ; 
from  (1)  and  (3),  x  =  8,  y  =  -I  ;  from  (2)  and  (3),  x  =  11, 
y  =  —  5.  From  this  it  is  evident  that  only  two  of  these 
equations  can  be  true  at  the  same  time.  Hence  the  problem 
is  impossible. 

If  (3)  had  not  been  independent,  but  deduced  from  (1) 
and  (2)  by  combining  them  in  any  way,  as,  for  example,  by 
adding  them,  giving 

8a  +  ty  =  29, 


DISCUSSION   OF  PROBLEMS.  213 

then  the  values  of  x  and  y  from  any  two  of  the  three  equations 
would  have  been  x  =  5,  y  =  7,  and  the  problem  would  have 
been  possible.  In  this  case  the  last  equation  would  have  been 
unnecessary.  When  a  problem  furnishes  more  conditions 
than  are  necessary  for  its  solution,  those  that  are  unnecessary 
are  termed  redundant. 

A  problem  which  involves  only  one  unknown  quantity  is 
sometimes  impossible.     Thus, 

What  number  is  that  whose  f|  increased  by  3  is  equal  to 
the  difference  between  its  |  and  ^  increased  by  11  ? 

Let  x  =  the  number  ;  then  we  have  the  equation 

y^  +  3  =  —  -  —  +  ii. 

15  6         10 
Clearing  of  fractions, 

22z  +  90  =  25a;  -  3x  +  330. 

.-.     0  =  240, 

which  shows  that  the  problem  is  absurd. 

Hence  it  appears  that  to  determine  any  number  of  unknown 
quantities,  there  must  be  given  as  many  independent  equa- 
tions as  there  are  unknown  quantities.  If  there  be  fewer, 
the  problem  is  indeterminate,  but  if  there  be  more,  the 
problem  in  general  is  impossible. 

103.  Discussion  of  Problems  —  Interpretation  of 
Negative  Results.  —  The  Discussion  of  a  Problem  consists 
in  assigning  different  values  to  the  arbitrary  quantities  which 
enter  the  equation,  and  interpreting  the  results. 

An  arbitrary  quantity  is  one  to  which  any  particular  value 
may  be  assigned  at  pleasure. 

It  sometimes  happens  in  the  solution  of  a  problem,  that 
the  result  obtained  has  the  minus  sign.  Such  a  result  is 
termed  a  Negative  Result ;  and  from  the  nature  of  positive 
and  negative  quantities  (Art.  20),  a  negative  result  must 
have  an  interpretation  diametrically  opposite  to  that  which 
it  would   receive   if    it   were   preceded    by   the  plus  sign. 


214  INTERPRETATION   OF  NEGATIVE  RESULTS. 

This  interpretation  we  will  illustrate  by  one  or  two  prob- 
lems. 

1.  What  number  must  be  added  to  a  in  order  that  the 
sum  may  be  b? 

Let  x  =  the  required  number  ;  then 
a  +  x  =  6. 

.-.     x  =  b  —  a, (1) 

which  gives  the  value  of  x  corresponding  to  any  assigned 
values  of  a  aud  b.  Thus,  if  a  =  12  and  b  =  25,  we  have 
from  (1), 

x  =  25  -  12  =  13. 

Here  x  is  +13,  which  means  that  13  must  be  added  to  12 
in  a  strict  Arithmetic  sense  in  order  to  make  the  sum  25. 

But  suppose  that  a  =  30  and  b  —  2-4  ;  then  (1)  becomes 
x  =  24  -  30  =  -6. 

What  is  the  meaning  of  this  negative  result?  The  problem 
now  reads  :  what  number  must  be  added  to  30  in  order  that 
the  sum  may  be  24  ?  It  is  evident  that,  if  the  words  added 
and  siim  are  to  retain  their  Arithmetic  meanings,  the  problem 
is  impossible.  If  however  the  problem  be  modified  so  as  to 
read  :  what  number  must  be  taken  from  30  in  order  that  the 
difference  may  be  24,  it  may  be  solved,  and  6,  which  is 
the  absolute  value  of  the  negative  result,  is  the  answer  to 
it.  This  second  problem  differs  from  the  given  problem  only 
in  this :  the  words  added  to  and  sum  are  replaced  b}'  the 
words  taken  from  and  difference. 

Now,  if  the  words,  number,  added  to,  and  sum,  be  used 
Algebraically,  then  the  negative  result,  —  G,  will  be  the 
correct  answer  to  the  given  problem.  In  this  case  the  nega- 
tive result  is  interpreted  as  meaning  that  —6  must  be  added 
Algebraically  to  30  to  produce  2-1,  or  that  G  must  be  sub- 
tracted Arithmetically  from  30  to  produce  24.  This  agrees 
exactly  with  the  principle  of  Art.  29,  that  subtracting  a 
positive  number  is  the  same  as  adding  an  equal  negative 
number. 


DISCUSSION   OF  PROBLEMS.  215 

2.    A's  age  is  35  years,  and  B's  age  is  20  years  :  when 
will  A  be  twice  as  old  as  B  ? 

Suppose  the  required  epoch  to  be  x  years  after  the  present 
time  ;  then  we  have 

35  +  x  =      40  +  2x (2) 


What  is  the  meaning  of  this  negative  result  ?  It  is  evident 
that  if  a  strictly  Arithmetic  meaning  is  to  be  given  to  the 
symbols  x  and  +,  equation  (2)  is  impossible,  for  40  is 
greater  than  35,  and  2x  is  greater  than  x,  so  that  the  two 
members  cannot  be  equal.  If  however  the  problem  be 
modified  so  as  to  read :  A's  age  is  35  years,  and  B's  age  is 
20  years  ;  when  was  A  twice  as  old  as  B,  it  may  be  solved, 
and  5,  which  is  the  absolute  value  of  the  negative  result,  is 
the  answer  to  it. 

Thus,  we  may  say  that,  in  the  given  example,  the  negative 
result  indicates  that  the  problem  in  a  strictly  Arithmetic 
sense  is  impossible,  but  that  a  new  problem  can  be  formed 
by  appropriate  changes  in  the  original  enunciation,  to  which 
the  absolute  value  of  the  negative  result  will  be  the  correct 
answer. 

Suppose  the  problem  had  originally  read  thus  :  A's  age  is 
35  years,  and  B's  age  is  20  years  ;  find  the  epoch  at  which 
A  is  twice  as  old  as  B.  Here,  if  we  suppose  that  the  required 
epoch  is  after  the  present  date,  we  obtain  for  the  required 
number  of  years  x  =  —5,  which  indicates  that  this  suppo- 
sition, in  an  Arithmetic  sense,  is  incorrect,  since  it  leads  to 
a  negative  result.  If  we  suppose  that  the  required  epoch  is 
before  the  present  date,  we  obtain  for  the  required  number 
of  years  x  =  5,  which  indicates  that  this  second  supposition 
is  correct,  since  it  leads  to  an  Arithmetic  value  for  x. 
Hence 

1.  We  may  say  that  a  negative  result  indicates  either  an 
erroneous  enunciation  of  a  problem,  or  a  wrong  choice  out  of 
two  p>ossible  suppositions  which  the  problem  allowed. 


216         INTERPRETATION   OF    THE   FORMS,    -,    — ,    -. 

0       CO       0 

2.  When  a  negative  result  is  obtained,  the  question  to  which 
it  is  the  answer  may  be  so  modified  as  to  make  the  result  true 
in  an  Arithmetical  sense. 

104.    Interpretation  of   the   Forms,  ?,  — ,  -.  —We 

0    c»    0 

shall  now  examine  the  meaning  of  certain  Forms  which 
occur  in  the  course  of  mathematical  operations,  and  Inter- 
pret them. 

The  symbol  0,  called  zero,  is  used  to  represent  either  abso- 
lute zero,  or  an  indefinitely  small  quantity,  i.e.,  a  quantity 
which  is  less  than  any  assignable  quantit}7.  If  we  subtract 
any  quantity  from  itself,  the  remainder  is  absolute  zero. 
Thus,  the  cipher  0  in  the  equation,  a  —  a  =  0,  is  the  abso- 
lute zero  of  Arithmetic. 

The  symbol  oo,  called  infinity,  is  used  to  represent  an 
infinite  quantity,  i.e.,  a  quantity  which  is  greater  than  any 
assignable  quantity. 

(1)  Let  -  represent  a  fraction.  If  the  numerator  a  re- 
mains constant  while  the  denominator  b  diminishes,  the  value 
of  the  fraction  increases.  If  the  numerator  remains  con- 
stant while  the  denominator  is  divided  by  any  number,  the 
value  of  the  fraction  is  multiplied  by  that  number  (Art.  85). 
When  the  denominator  becomes  veiy  small  the  value  of  the 
fraction  becomes  very  large  ;  and  finally,  when  the  denomi- 
nator becomes  less  than  any  assignable  quantity,  or  0,  the 
value  of  the  fraction  becomes  greater  than  any  assignable 
quantity,  or  infinity.     That  is, 

-  =  oo (1) 

0  V 

This  may  be  otherwise  expressed  as  follows  :  If  a  fraction 
has  a  fixed  numerator,  no  matter  how  small,  the  denominator 
can  be  made  so  much  smaller  than  the  numerator,  that  the 
fraction  shall  be  greater  than  any  assignable  quantity. 

(2)  If    the  numerator  a  remains  constant  while  the  de- 


INTERPRETATION   OF    THE   FORMS,    -,    — ,    -.         217 

0      cc       0 

nominator  b  increases,  the  value  of  the  fraction  decreases 
(Art.  86).  When  the  denominator  becomes  very  large  the 
value  of  the  fraction  becomes  very  small ;  and  finally,  when 
the  denominator  becomes  greater  than  any  assignable  quan- 
tity, or  co,  the  value  of  the  fraction  becomes  less  than  any 
assignable  quantity,  or  0.     That  is, 

2l  =  0 (2) 

00 

This  may  be  otherwise  expressed  as  follows  :  If  a  fraction 
has  a  fixed  numerator,  no  matter  how  large,  the  denominator 
can  be  made  so  much  larger  than  the  numerator,  that  the 
fraction  shall  be  less  than  any  assignable  quantity. 

The  zero  0  in  (1)  and  (2)  is  an  abbreviation  to  denote  an 
indefinitely  small  quantity,  and  does  not  mean  absolute  zero. 

If  a  =  0  and  b  is  not  =0,  then  of  course  the  value  of  the 
fraction  =0. 

(3)    If  a  =  0  and  6  =  0,  the  value  of  the  fraction  -  takes 

the  form  g.  In  this  case,  the  fraction  may  be  said  to  have 
any  value  whatever;  that  is,  since  the  divisor  zero,  multiplied 
by  any  number  whatever,  gives  the  dividend  zero,  therefore 
zero,  divided  by  zero,  may  have  any  value  whatever.  Hence 
the  fraction  J}  is  indeterminate,  and  is  called  the  Symbol  of 
Indetermination.  * 

This  symbol,  however,  does  not  always  mean  indetermi- 
nation; but  is  often  the  result  of  a  particular  supposition, 
which  makes  a  factor  become  zero,  which  is  common  to  both 
terms  of  a  fraction.     Thus  : 

1    Let  it  be  required  to  find  the  value  of  the  expression 

x2  —  a"1     , 

when  x  =  a. 


If  we  suppose  x  =  a,  this  becomes  §,  the  value  of  which 
is  indeterminate.     But   if   we   cancel   the   common    factor, 

*  Called  also  a  Vanishing  Fraction. 


218  PROBLEM  OF  THE    COURIERS. 

x  —  a,  before  we  put  x  =  a,  this  becomes  x  +  a.  If  we 
now  suppose  x  =  a,  the  expression  becomes  2a,  which  is  its 
true  value. 

Here  we  see  that  the  expression  becomes  °-  in  consequence 
of  the  factor  x  —  a  appearing  in  both  numerator  and  de- 
nominator. Now  we  cannot  divide  by  a  zero  factor,  but  as 
long  as  x  is  not  absolutely  equal  to  a  the  factor  x  —  a  may 
be  removed. 

2     Find  the  value  of  the  fraction 

or  —  a2       , 

,  when  x  =  a. 


(x  -  ay 


If  we  suppose  x  =  «,  this  expression  becomes  J).  But 
if  we  first  cancel  the  common  factor,  x  —  a,  and  then  put 

x  =  a,  the  expression  becomes  —  =  <x>. 

0 

3     Find  the  value  of  the  fraction 

(x  —  a)'2      , 

^- '—,  when  x  =  a. 

x-  —  <r 

If  we  suppose  x  =  a,  this  expression  becomes  •{].  But  if 
we  first  cancel  the  common  factor,  x  —  a,  and  then  put  .'•  =  a, 

the  expression  becomes  —  =  0. 
1  2a 

These  examples  show  that  the  true  value  of  a  fraction. 
which  reduces  to  the  form  {J,  may  be  either  finite,  infinite,  or 
zero,  and  therefore,  before  we  decide  that  it  is  really  indeter- 
minate, we  must  ascertain  whether  the  apparent  indetermina- 
tion  has  not  arisen  from  the  existence  of  a  common  factor, 
which  becomes  <•  under  a  particular  supposition. 

105.  Problem  of  the  Couriers. *  — Two  couriers,  A 
and  B,  travel  along  the  same  road,  and  in  the  same  direction, 
It'Ii,  at  the  rates  of  m  and  n  miles  per  hour  respectively. 
At   a   certain   time,  say  12   o'clock,  A  arrives   at   the  point  P 

*  Originally  proposed  by  Clalraut. 


PROBLEM    OF   THE   COURIERS.  219 

and  B  arrives  at  the  point  Q  distant  a  miles  from  P :    find 
when  and  where  they  will  be  together. 

^_ p __£> R 

The  discussion  of  this  problem  will  serve  to  further  illus- 
trate the  principles  of  Arts.  103  and  104,  and  to  show  that 
the  results  of  every  correct  solution  correspond  to  the  cir- 
cumstances of  the  problem. 

Let  the  point  P  be  taken  as  the  origin  of  distances,  and 
suppose  that  distances  measured  to  the  right  from  P,  and 
time  counted  after  12  o'clock,  are  regarded  as  jwsitive. 

Let         t  =  the  number  of  hours  from  12  o'clock  to  the 
time  of  meeting, 
and  x  =  the  distance  A  travels  in  t  hours  ; 

then  x  —  a  =  the  distance  B  travels  in  t  hours. 

Now,  since  the  distance  traveled  by  each  equals  the  rate 
per  hour  multiplied  by  the  number  of  hours,  we  have 
x  =  mt,     and     x  —  a  =  nt. 

Solving  these  equations  for  t  and  x,  we  get 

«--2- (i) 

m  —  n 

ma                        .  ,as 

*  =   ■ (2) 

m  —  n 

We  shall  now  examine  these  values  of  t  and  x  on  different 
suppositions  as  to  the  arbitrary  quantities,  a,  m,  and  n. 

1.  Suppose  m  >  n,  and  a  >  0. 

This  hypothesis  makes  both  terms  of  the  fractions  (1)  and 
(2)  positive  ;  hence  the  values  of  both  t  and  x  are  positive. 
That  is,  the  couriers  are  together  after  12  o'clock,  and  to  the 
right  of  P. 

This  interpretation  evidently  corresponds  with  the  supposi- 
tion made.  For  if  m  >  n,  A  travels  faster  than  B,  and  will 
therefore  be  continually  gaining  on  him  ;  hence  at  some  time 
after  12  o'clock  they  must  be  together  and  at  the  right  of  P. 

2.  Suppose  m  <  n,  and  a  >  0. 

This  hypothesis  makes  the  numerator  in  each  of  the 
fractions   in  (1)   and   (2)    positive,    and   the   denominators 


220  PROBLEM   OF   THE    COURIERS. 

negative  ;  hence  the  values  of  both  t  and  x  are  negative. 
Now,  since  we  supposed  distances  measured  to  the  right  of 
P,  and  time  counted  after  12  o'clock  to  be  regarded  as 
positive,  we  conclude,  from  what  we  have  observed  respect- 
ing negative  quantities  (Arts.  20,  29,  103),  that  these  values 
of  t  and  x  indicate  that,  instead  of  the  couriers  meeting  after 
12  o'clock  and  to  the  right  of  P,  they  met  before  12  o'clock 
and  to  the  left  of  P. 

This  interpretation  evidently  corresponds  with  the  sup- 
position made.  For,  if  m  <  n,  A  travels  more  slowly  than 
B,  and  hence,  after  12  o'clock,  they  continually  separate. 
But  as  A  travels  more  slowly  than  B,  and  was  behind  B  at 
12  o'clock,  it  follows  that,  at  some  time  before  12  o'clock 
and  at  the  left  of  P,  they  must  have  been  together. 

3.  Suppose  vi  =  7i,  and  a  >  0. 

This  hypothesis  makes  the  numerator  of  each  fraction 
finite  and  the  denominator  zero ;  hence,  the  values  of  both  t 
and  x  become  <x>  (Art.  104).  That  is,  the  couriers  will  be 
together  at  an  infinite  distance  from  P;  in  other  words,  they 
will  never  be  together. 

This  interpretation  evidently  corresponds  with  the  sup- 
position made.  For,  if  m  =  n,  the  couriers  travel  at  the 
same  rate  ;  and  since  they  were  a  miles  apart  at  12  o'clock, 
they  always  have  been,  and  always  will  be  separated  by  the 
same  distance,  so  that  they  will  never  meet. 

4.  Suppose  m  >  n  or  <  n,  and  a  —  0. 

This  hypothesis  makes  the  numerator  of  each  fraction 
zero,  and  the  denominator  finite  ;  hence  the  values  of  t  and 
a;  become  0  (Art.  104).  That  is,  the  couriers  are  together 
at  the  point  P  at  12  o'clock,  and  at  no  other  time  and  place. 

This  interpretation  evidently  corresponds  with  the  sup- 
position made.  For,  if  a  =  0,  the  couriers  are  together  :it 
12  o'clock  at  the  point  P;  and  as  m  and  n  arc  unequal,  they 
travel  at  different  rates  ;  hence  they  can  never  be  together 
after  12  o'clock,  nor  could  they  ever  have  been  together  be- 
fore that  time. 


EXAMPLES.  221 

5.    Suppose  m  =  n,  and  a  =  0. 

This  hypothesis  makes  both  terms  of  each  fraction  0 ; 
hence  the  values  of  both  t  and  x  assume  the  form  -g.  But 
when  an  expression  takes  this  form,  it  may  have  any  value 
whatever  (Art.  104(3)).  That  is,  there  is  an  infinite  number 
of  times  when  the  couriers  are  together. 

This  interpretation  evidently  corresponds  with  the  sup- 
position made.  For,  if  m  =  n  and  a  =  0,  the  couriers  were 
together  at  12  o'clock  at  the  point  P,  and  were  traveling  at 
the  same  rate  ;  hence  they  are  always  together,  both  before 
and  after  12  o'clock. 

EXAMPLES. 

1.  Given   2x  —    y  =     2,  ox  —  dy  =    3, 

ox  +  2y  =  17,  4x  +  dy  =  24  ; 

to   find   x   and   y,   and   to    show  how  many  equations   are 
redundant  (Art.  102).  Ans.  x  =  3,  y  =  4. 

2.  The  sum  of  two  numbers  is  9,  and  their  difference  is 
25:  find  the  numbers.  Ans.  17,  and  —8. 

Interpret  the  negative  result  obtained,  and  modify  the 
question  accordingly  (Art.  103). 

3.  A  is  50  years  old,  and  B  40  :  find  the  time  when  A  will 
be  twice  as  old  as  B.  Ans.  In  —30  years. 

Interpret  the  negative  result,  and  modify  the  question 
accordingly. 


4, 

T^iihI  flip  vnlup  nf*        *                        wlion  t  —    1 

Ans.  %. 

X    ill  11     Lll^      >  ttlLLO    KJL         „                                                 VV  11^11    •*/     —      x« 

x2  +  x  —  2 

5. 

Find  the  value  of  — when  x  =  a. 

x-  —  a2 

la. 

6. 

Find  the  value  of  — when  x  =  a. 

x'2  —  ax 

2. 

7. 

Find  the  value  of  ■ — :r— when  x  = 

a.            3. 

222  JNE  Q  UALITIES. 

INEQUALITIES. 

106.  Inequalities.  —  It  is  often  desirable  to  compare 
expressions  that  are  unequal,  and  to  learn  which  of  them  is 
the  greater. 

An  Inequality  is  a  statement  in  Algebraic  language  that 
two  expressions  are  unequal.     Thus, 

a  >  &, 

is  an  inequality,  showing  that  a  is  greater  than  &,  or  that 
a  —  b  is  positive.     Also, 

a  <  b, 

is  an  inequality,  showing  that  a  is  less  than  b,  or  that  a  —  b 
is  negative. 

The  quantity  on  the  left  of  the  sign  is  called  the  first 
member,  and  that  on  the  right,  the  second  member  of  the 
inequality. 

The  form  a>  b  >  c, 

indicates  that  b  is  less  than  a  but  greater  than  c. 

The  form  a  >  b, 

indicates  that  a  may  be  cither  equal  to  or  greater  than  />,  but 
cannot  be  less  than  b. 

Two  inequalities  are  said  to  subsist  in  the  same  sense  when 
the  first  member  is  the  greater  or  the  less  in  both.  Thus,  the 
inequalities 

a  >  &,     and     5  >  3  ;     or     a  <  b,     and    3  <  8, 
are  said  to  subsist  in  the  same  sense. 

Two  inequalities  are  said  to  subsist  in  a  contrary  sense 
when  the  first  member  is  the  greater  in  one,  and  the  less  in 
the  other.     Thus,  the  inequalities 

a  >  b,     and     3  <  8  ;     or     a  <  b,     and     7  >  2, 
are  said  to  subsist  in  a  contrary  sense. 

Properties  of  Inequalities.  —  (1)  An  inequality  will  still 
subsist  in  the  same  sense  after  the  same  quantity  has  been 
added  to  or  subtracted  from  each  member. 


INEQUALITIES.  223 

For  suppose  a  >  b,  then  a  —  b  is  positive  ;  therefore, 

a  +  c  —  (6  +  '')     and     «  —  c  —  (b  —  c) 

are  positive,  since  each  equals  a  —  b.     Therefore 

a  -f-  c  >  b  +  c,     and     a  —  c>  6  —  c. 

Hence,  a?i?/  ienw.  may  6e  transposed  from  one  member  of 
an  inequality  to  the  other,  if  its  sign  be  changed. 

(2)  If  the  signs  of  all  the  terms  of  an  inequality  be  changed, 
the  sign  of  inequality  must  be  reversed. 

For,  to  change  all  the  signs  is  equivalent  to  removing  each 
term  of  the  first  member  to  the  second,  and  each  term  of  the 
second  member  to  the  first. 

(3)  An  inequality  will  still  subsist  in  the  same  sense  after 
each  member  has  been  multiplied  or  divided  by  the  same 
positive  number. 

For  suppose  a  >  b,  then  a  —  b  is  positive  ;  therefore,  if  m 
is  positive 

m(a  —  b)     or     —(a  —  b) 
m 

is  positive  ;  and  therefore 

,  a      b 

ma  >  mb     or     —  >— . 
m     7/1 

It  may  be  shown  in  like  manner  that  if  each  member  of 
an  inequality  be  multiplied  or  divided  by  the  same  negative 
number,  the  sign  of  inequality  must  be  reversed. 

(4)  If  tivo  or  more  inequalities  subsist  in  the  same  sense, 
the  corresponding  members  may  be  added  together,  and  the 
resulting  inequality  will  subsist  in  the  same  sense. 

For,  if  a  >  b,  a'  >  b' ',  a"  >  &",  ......  then  a  —  b,  ar  —  b' , 

a"  —   b" , are   all   positive ;    therefore   their   sum 

a  —  &  +  a!  —  .&'  +  a"  —  b"  + is  positive ;  hence 

a  +  a'  +  a"  +  .   .   .   .  >  b  +  b'  +  b"  + 

(5)  If  a>b,  a'  >  b',  a"  >  b" , and  all  the  quantities 

are  positive,  then  it  is  evident  that  aa'a" >  bb'b"  .... 


224  EXAMPLES   OF  INEQUALITIES. 

(G)  If  tioo  positive  members  of  an  inequality  be  raised  to 
any  power,  the  inequality  will  still  subsist  in  the  same  sense. 

For  suppose  a  >  6,  and  let  a  and  b  be  positive  ;  then  if  n 
is  any  positive  number,  it  follows  directly   from   (5)  that 

a"  >  b". 

EXAMPLES    OF    INEQUALITIES. 

1.  Find  the  limit  of  x  in  the  inequality 

Clearing  of  fractions, 

GO  +  4a;  >  9(5  +  3.x-. 
x  >  36. 

2.  If  a  >  b,  prove  that  a-  +  b"  >  2ab. 

The  square  of  (a  —  b)  or  (6  -  a)  must  be  positive  (Art. 
36),  and  therefore  greater  than  zero. 

...     r,2  _  2db  4-  &2>0. 
'   .-.     a2  4-  &2>  2a&. 

a.  4-  tt„  4-  cl  4-  .  .  .  .  a„      ,, 

3.  Show  that  the  fraction  -     ,     *     ,     *  ~ f  >  the 

\  +  62  4-  68  4-  .  .  .  .  &„ 

least,  and  <  the  greatest  of  the  fractions,  %  \  f'm''  ' '/','' 
all  of  the  denominators  having  the  same  sign. 

Suppose  the  fractions  %  ^2,  f\  .  .  .  .  ','"  to  be  in  ascend- 

°1      b-l      }):\  h" 

ing  magnitude,  and  that  all  the  denominators  are  positive. 
Then 


...     a,  =  \  X  "b 

«2    >    <hy 

k    b; 

.-.     aa  >  &a  x  ^b 

2a  >  "> 

.-.     a,  >  l>..  x  "b 
6, 

and  so  on. 

EXAMPLES.  225 


Adding,  a1+a2+as+ an>  (&1+&2+68+ .  .  .  .  hn)j 


(h  +  °2  +  <ls  +  •  •  •  •  rf»^  ^1 

&!  +   &2   +  .&8   + &»  "  &l' 

Similarly  it  may  be  shown  that 

ft,  +  «a  +  «»+••«.  ft, 


< 


bx  +  &2  +  ^3  +  •  .  .  .  &„       &„ 

In  like  manner  the  proposition  may  be  proved  when  all  the 
denominators  are  negative. 

Find  the  limit  of  x  in  the  following  three  inequalities  : 

Ans.  x  >  5. 
x>2. 


4. 

7a:  —  3  >  32. 

5. 

7jb  -  %3  >  fa;  +  5. 

6. 

ax   ,,            r     ft2 
5                           5 

7. 

Show  that  -  -+-  -  >  2,  unless  a  =  b. 
b       a 

8. 

Show  that  w3  +  1  >  ?r  +  w,  unless 

EXAMPLES. 

1.    Show  that  the  following  example  is  impossible : 
What  number  is  that  whose  T\  diminished  by  5  is  equal  to 
the  difference  between  its  f  and  £  increased  by  7  ? 

Interpret  the  negative  results  in  the   following  two  ex- 
amples, and  modify  the  enunciation  accordingly. 

2.  The  triple  of  a  certain  number  diminished  by  100  is 
equal  to  4  times  the  number  increased  by  200 :  find  the 
number.  Ans.  -300. 

3.  The  sum  of  two  numbers  is  00,  and  their  difference 
is  120:  find  the  numbers.  Ans.  105,  —15. 

x3  —  1 

4.  Find  the  value  of  — — ,  when  a;  =  1. 

a;3  —  2x2  +  2x  —  1 

Ans.  3. 

5.  Find  the  value  of        ~     ,  when  x  =  1.  5. 


226 

6.  Find  the  value  of 

7.  Find  the  value  of 

8.  Find  the  value  of 

(4a;3  -  9)(1  +  x) 

Ans.  %. 
Find  the  limit  of  x  in  the  following  four  inequalities : 

9.  3x  —  5  >  13.  Ans.  x  >  6. 

10.    I^_^>*-3. 
5         3      3 


EXAMPLES. 

&  _  5^  +  7.v  _  3      i 

c  =  3. 

ar  —  a-  —  5a;  —  3 

Ans.  \ 

rz  4-  1 

* — — — ,  when  x  =  —  1. 

ar  —  1 

_3 
3 

(3  + 2a*)  (.  -5) 

X  >  0. 

p  —  m 

q  —  ri 

x  >  a. 

11.  m  —  nx  >  p  —  gas. 

12.  ax  +  5&a;  —  5a&  >  a2. 

In  the  following  examples  the  letters  are  unequal   and 
positive. 

b      1    ,    1 

ijt-,>-  +  T- 
b"       cr    a       b 

14.  Show  that  a"  +  b'2  +  c2  >  o/>  +  oc  +  &c. 

15.  Show  that  (a2  +  &2)(a4  +  &4)  >  (a8  +  &3)2. 

16.  If  x*=  a2+&2,  and  y2  =  c2  +  d2,  show  that  xy  >  ac  +  bd. 

17.  Show  that  abc  >  (a  +  b  —  c)(a  +  c  —  b)  (6  +  c  —  0). 

18.  Show  that  «b(a  +  6)  +  be  {b+c)  +  w'(''  +  a)  >  Gabc. 

19.  Show  that  (a  +  &)(&  -I-  c){c  +  a)  >  8a6c. 

20.  Show  that  x2  —  8x  +  22  is  never  less  than  6,  what- 
ever may  be  the  value  of  x. 

21.  Show  that  a  +  c  +  e  >  the  least,  and  <  the  greatest 

b  +  d  +  f 

-,   -,    -,  each  letter  being  positive. 
b    d    f  m 


INVOLUTION    OF  POWERS    OF  MONOMIALS.  227 


CHAPTER     XII. 

INVOLUTION     AND     EVOLUTION. 

107.  Involution  is  the  process  of  raising  an  expression 
to  any  required  power.  Involution  is  therefore  only  a 
particular  case  of  multiplication,  in  which  the  factors  are 
equal  (Art.  12).  It  is  convenient,  however,  to  give  some 
rules  for  writing  clown  the  power  at  once. 

It  is  evident  from  the  Rule  of  Signs  (Art.  36)  that, 

(1)  No  even  power  of  any  quantity  can  be  negative. 

(2)  Any  odd  power  of  a  quantity  will  have  the  same  sign 
as  the  quantity  itself.     Thus, 

(  —  a)2  =  (  —  «)(  — a)  =  +a~, 
(  —  ay  =  (  —  a)  (  —  a)  (  —  a)  =  +a2(  —  a)  =  —a3, 
{-ay  =  (-«)(-«)  (-a)  (-«)  =  (-a*)  (-a)  =  +a4; 
and  so  on. 

Note. —  The  square  of  every  expression,  whether  positive  or 
negative,  is  positive. 

108.  Involution  of  Powers  of  Monomials.  —  From 
the  definition,  we  have,  by  the  rules  of  multiplication, 

(a2)8=  (a2)  (a2)  (a2)  =  o2+2+2  =  a6. 
(-a8)2=(-a8)(-a8)=a8+8  =  a6. 
(_3a3)2=(-3a3)(-3a3)  =  (-  3)2(a3)2  =  9a6. 
Generally, 

(«,"')"= «m  •  am  •  am  •  am  .  .  .  .  to  n  factors 

__am  +  m  +  m+m fa  n   termg 

=  a""4. 

(ab)m=ab  •  ab  •  ab to  m  factors 

=  (aaaa  ....  to m  factors)  x  (bbbb  ....  torn  factors) 
=ambm. 

Hence  (ab)"1  =  a'"b"\ 


228 


EXAMPLES. 


and  so  on  for  any  number  of  factors.  Thus,  the  mth  poiver 
of  a  product  is  equal  to  the  product  of  the  m'h  powers  of  its 
factors. 

Hence  (a/W )"<  =  {ax)m(p»)m(<f)m  ...... 

=  asanbvmczm 

Hence,  to  raise  any  power  of  a  quantity  to  any  other 
power,  we  have  the  following 

Rule. 

Raise  the  numerical  coefficient  to  the  required  power  by 
Arithmetic,  multiply  the  exponent  of  each  factor  by  the 
exponent  of  the  required  power,  and  give  the  proper  sign 
to  the  result. 


1) 

n      a       a       a                        ~    . 

=  -  X       X  - torn  factoi 

b        b        b 

aaa  ....  to  m  factors 

bbb  .  ...  torn  factors 

_  am 

Hence,  for  obtaining  any  power  of  a  fraction  :  Raise  both 
the  numerator  and  denominator  to  that  power,  and  give  the 
proper  sign  to  the  result. 


EXAMPLES 
Show  that 

1.  (a268)2=  a466. 

2.  (-a2b3)s=  -«r'b». 

3.  (aW)4  =  aWV6. 

4.  (— 2<c2)6  =  -32s10. 

5.  (_3a&8)6  =  729a6&18. 


\lr)         b« 
\     be*}  b:>cu 


8.  /-JLV  =_«!_, 

\    b-cy      612c18 

11.    /     .H/V\-=  .rV 


27a;16 
125**' 


INVOLUTION   OF  BINOMIALS.  229 

109.  Involution  of  Binomials.  —  We  have  already 
proved  by  actual  multiplication  (Art.  41),  the  two  following 
cases  of  the  involution  of  binomial  expressions  : 

(a  +  b)2  =  a-  +  2ab  +  &2    .     .     .     .   (1) 

(a  -  b)2  =  a2  —  2ab  +  62    .     .     .     .   (2) 

If  we  multiply  (1)  and  (2)  by  a  -{-  b  and  a  —  6  respec- 
tively, we  have 

(a  +  6)3  =  a3  -(-  3a26  +  3a&2  +  b3  .     .  .   (3) 

(a  -  b)s  =  a3  -  3a26  +  3a&2  -  &3 .     .  .   (4) 

If  we  multiply  (3)  and  (4)  by  a  +  b  and  a  —  b  respec- 
tively, we  shall  have 

(a  +  6)4  =  a4  +  4a3&  +  Ga2b2  +  4ab3  +  b\ 

(a  -  by  =  a4  -  4a86  +  6a2&2  -  4a&8  +  b\ 

By  multiplying  these  two  results  by   a  -f-  &  and  a  —  b 

respectively  we  should  obtain  (a  +  &)5  and  (a  —  &)5;   and 

by   continuing   the   process  we    could   obtain   any  required 

power  of  (a  +  b)  or  (a  —  b).     Hence  the  following 

Rule. 

Multiply  the  binomial  by  itself,  until  it  has  been  taken  as  a 
factor  as  many  times  as  there  are  units  in  the  exponent  of  the 
required  poiver. 

This  rule,  however,  would  be  very  laborious  in  finding  any 
high  power,  for  instance  (a  +  6)20.  In  Chapter  XIX.  we 
shall  prove  a  theorem,  called  the  Binomial  Theorem,  by  the 
aid  of  which  any  power  of  a  binomial  expression  can  be 
obtained  without  the  labor  of  actual  multiplication. 

Since  the  above  formulae  are  true  for  all  values  of  a  and  b, 
we  can  write  down  the  squares  and  the  cubes  of  any  binomial 
expressions.     Thus, 

1.    (a*  -  V)2  =  (a4)2  +  2«K-//)  +  (-b*)2 

=  a8  -  2a%*  +  bs. 


230  INVOLUTION   OF  POLYNOMIALS. 

Show  that 

2.  (2x  +  3y)2  =  4a-2  +  12.r?/  +  9?/2. 

3.  (3a;  +  5?/)2  ==  9a-2  +  SOxy  +  25?/2. 

4.  (a;  -  2#)3  =  a;3  -  6afy  +  12av/2  -  Sys. 

5.  (2a6  -  3c)3  =  8a3Z>3  -  36a262c  +  Mabc2  -  27c3. 

G.    (5a2  -  362)3  =  125a°  -  225a462  +  135a?b4  -  2766. 

110.    Involution   of    Polynomials.  —  We  may   now 

apply  the  formulae  of  Art.  109  to  obtain  the  powers  of  any 
trinomial  or  polynomial.     Thus  from  (1) 

(«  +  &  +  c)-=  [(a  +  b)  +  c]2 

=  (a  +  &)2  +  2(a  4-  &)c  +  c2 

=  a2  +  &2  +  c2  +  2ab  +  2ac  +  26c     .   (1) 

In  the  same  way  we  may  prove 

(a+b+c-\-d)2=a-+b2+c-+d2+2ab+2ac+2ad+2bc+2bd+2rd    .    (2) 

We  observe  in  both  (1)  and  (2)  that  the  square  consists  of 

(1)  the  sum  of  the  squares  of  the  several  terms  of  the 
given  expressions  ; 

(2)  twice  the  sum  of  the  products  two  and  two  of  the 
several  terms,  taken  with  their  proper  signs. 

The  same  law  holds  whatever  be  the  number  of  terms  in. 
the  expression  to  be  squared.     Hence  the  following 

Rule. 

To  find  the  square  of  any  polynomial,  write  the  square  of 
each  term  together  with  twice  the  product  of  each  term  by  each 
of  the  terms  following  it. 

From  (3)  of  Art.  109  we  obtain  the  cube  of  a  trinomial  as 
follows  : 
(a+b+cf  =  [0+(6+c)]« 

-  a»+Saa(&+c)+8a(&+c)*+(&+c)" 

—  a8+68+c8+3aa(6+c)+868(o+c)+8ca(a+5)+6oBc     .  (8) 


INVOLUTION    OF  POLYNOMIALS.  231 

Hence  to  find  the  cube  of  a  trinomial  we  have  the  following 

Rule. 

Write  the  cube  of  each  term,  together  with  three  times  the 
product  of  the  square  of  each  term  by  the  sum  of  the  other 
two,  and  six  times  the  product  of  the  three  terms. 

Formulae  (1),  (2),  and  (3)  may  be  used  for  obtaining  the 
squares  and  cubes  of  any  polynomial  expressions,  as  ex- 
plained in  Art.  109.  Thus,  if  we  require  (1  —  2x  +  3a;2)2, 
iu  formula  (1)  we  put  1  for  a,  —2x  for  b,  and  3a:2  for  c,  and 
obtain 

1.  (l-2x+3x2y 

=  (l)2+(-2x-)2+(3^)2+2(l)(-2.i-)+2(l)(3.r)+2(-2a;)(3^) 
=  1 + 4a:2 + 9a;4-4a;+  Gar- 1 2a:3 
=  l-4a;+10a;2-12a;3+9a:4. 

Similarly  by  (3)  we  have 

2.  (1-2Z+3Z2)3 

=  (l)3+(_2^)3+(3a;2)3-f3(l)2(-2.T+3.«2)  +  3(-2a;)2(l+3a;2) 
+3(3.x-2)2(l-2a;)+6(l)(-2a;)(3a;2) 

=  l_8x3+27a;6+3(-2.T+3x-2)  +  12a;2(l+3^) 

+  27a;4  (1 -2a;) -36a;3 
=  l_Ga;+21a;2-44a;3+63a;4-54a;5+27a;6. 

Show  that 

3.  (i  _  x  _|_  x2)2  „  :  _  2x  +  3a;2  -  2.x-3  +  %*• 

4.  (1  +  3a;+2a;2)2  =  1  +  6a;  +  13a;2  +  12a;3  +  4a;4. 

6.  (fa-2  -  x  +  f)2  =  ^  -  ^  +  3a;2  -  3a;  +  f . 

7.  (1  +  x  +  a;2)3  =  1  +  3a;  +  6a:2  +  7a-3  +  Car4  +  3a-5  +  a6. 

8.  (1  +  a;  -  a;2)3  =  1  +  3a  -  5a;3  +•  3a;5  -  a;6. 


232  EVOLUTION   OF   MONOMIALS. 

EVOLUTION. 

11L  Evolution  —  Evolution  of  Monomials.  —  Evo- 
lution is  the  operation  of  finding  any  required  root  of  a 
number  or  expression.  A  root  of  any  quantity  is  a  factor 
which  being  multiplied  by  itself  a  certain  number  of  times 
produces  the  given  quantity  (Art.  13).  Hence  Evolution  is 
the  inverse  of  Involution  (Art.  107). 

The  symbol  which  denotes  that  a  square  root  is  to  be 
extracted  is  \j~ ;  and  for  other  roots  the  same  symbol  is 
used,  but  with  a  figure  called  the  index  written  above  to 
indicate  the  root  (Art.  13). 

By  the  Rule  of  Signs  (Art.  36),  we  see  that 

(1)  any  even  root  of  a  positive  quantity  may  be  either 
positive  or  negative; 

(2)  every  odd  root  of  a  quantity  has  the  same  sign  as  the 
quantity  ; 

(3)  there  can  be  no  even  root  of  a  negative  quantity. 
Thus,  (l)a  x  a  =  a2,  and  (— a)(— a)=  a2;  therefore  there 
are  two  roots  of  a2,  namely,  -j-a  and  —a. 

(2)  (  —  «)(  — a)  (  —  a)  =  —  a3;  therefore  the  cube  root 
of   —  a3  is  —a. 

(3)  There  can  be  no  square  root  of  —a'2;  for  if  any 
quantity  be  multiplied  by  itself,  the  result  is  a  positive 
quantity. 

There  can  be  no  even  root  of  a  negative  quantity,  because 
no  quantity  raised  to  an  even  power  can  produce  a  negative 
result.  Even  roots  are  called  impossible  roots  or  imaginary 
roots. 

Since  the  n"'  power  of  a'"  is  am"  (Art.  108),  it  follows  that 
the  /'"'  root  of  a'""  is  a"1. 

Also,  the  mih  power  of  a  product  is  the  product  of  the  m'h 
powers  of  its  factors  (Art.  108)  ;  hence,  conversely,  the 
mth  root  of  a  product  is  the  product  of  the  m"'  roots  of  its 
factors.     Thus, 

yJaTc  =  \Ja  sjb  V7'7;  "ylab  =  7«  V&. 


EVOLUTION   OF  MONOMIALS.  233 

Again,  we  have  (Art.  108) 


(axbycr  ....)' 


therefore,  conversely,  mJaxmV/m<fm  .  .  .  .  =  axb"c? 


Hence  to  extract  any  root  of  a  monomial,  we  have  the 
following 

Rule. 

Extract  the  required  root  of  the  coefficient  by  Arithmetic, 
then  divide  the  exponent  of  every  factor  in  the  expression  by 
the  index  of  the  root,  and  give  the  proper  sign  to  the  result. 

Thus,  for  example, 

s/a1  =  a2 ;     V7^4  =  a862 ;     v^  =  -x3 ;     ffi  =  x2 ; 


/lGa-64  =  lab2;     r/-8aW2  =  -2aW. 

To  obtain  any  root  of  a  fraction  :  Find  the  root  of  the 
numerator  and  denominator,  and  give  the  proper  sign  to 
the  result. 

This  is  the  converse  of  the  rule  in  Art.  108. 


,     4s/-27a6  3a2 

For  example,  ^-^  =   ~4y 

Note  1.  — Since  every  positive  quantity  has  two  square  roots  equal 
in  magnitude  but  opposite  in  sign,  it  is  customary  to  prefix  the  double 
sign  ±,  read  plus  or  minus,  to  a  quantity  when  we  wish  to  indicate 
that  it  is  either  +  or  — .     Thus 


?/256a;y  =  (/iV/  =   ±ixy\ 

Note  2.  —  Any  quantity  whose  root  can  be  extracted  is  called  a 
perfect  power.  "When  the  square  root  of  an  expression  which  is  not 
a  perfect  square,  or  the  cube  root  of  an  expression  which  is  not  a 
perfect  cube,  is  required,  the  operation  cannot  be  performed.  Thus 
we  cannot  take  the  cube  root  of  a-  since  the  exponent  2  is  not  divisible 
by  the  index  3.  At  present  we  can  only  express  the  result  thus  ?/  d1. 
Also,  \fa,  \[az,  \fab,  cannot  at  present  be  otherwise  expressed  ;  and 
similarly  in  other  cases.  Such  quantities  are  called  surds  or  irrational 
quantities,  and  will  be  considered  in  Chapter  XIII. 


234  SQUARE  ROOT   OF  A   POLYNOMIAL. 

EXAMPLES. 
Show  that 


1.    s/oW2  =  ±aibd 


6  . 


2.    ^G4xYs  =   ±Sx3f; 


,       /28U//V 
'   V    81a;10 

4.   V27oW  =  3a26c3 


9a;5 


5.  ^_343a126u  =  -7a*bG; 

6.  *0^=_3s> 

7.  V^y1*7  =  .«yz ; 

8.  V-^y 


112.  Square  Root  of  a  Polynomial.  —  Since  the 
square  of  a  +  &  is  a2  +  2«6  +  &2,  the  square  root  of 
a2  -f-  2a,b  -f-  62  is  a  -f-  6.  "We  may  deduce  the  general  rule 
for  the  extraction  of  the  square  root  of  a  polynomial  by 
observing  in  what  manner  a  +  b  may  be  derived  Iron: 
a2  +  2ab  +  b2. 

Arrange  the  terms  of  the  square  according  to  the  descend- 
ing powers   of   a ;    then  the  first 
term  is  a2,  and  its  square  root  is       a    +  2«&  +  b2  |a  +  b 

a,  which  is  the  first  term  of  the       &_ 

required  root.     Subtract  its  square,    2a  +  b)2ab  +  6a 
a2,  from  the  given  expression,  and  2ab  +  &2 

bring  down  the  remainder,  2ab  -+-  b2. 

Thus,  b,  the  second  term  of  the  root,  will  be  the  quotient 
when  2ab,  the  first  term  of  the  remainder,  is  divided  by  2a, 
i.e.,  by  double  the  first  term  of  the  root.  This  second  term, 
6,  added  to  2a,  twice  the  first  term,  completes  the  divisor, 
2a  +  b  ;  multiply  this  complete  divisor  by  b,  the  second 
term,  and  subtract  the  product,  i.e.,  2ab  +  b2,  from  the 
remainder,  and  the  operation  is  completed. 

If  there  were  more  terras  we  should  proceed  with  a  +  b 
as  we  have  done  with  a ;  its  square,  a2  -f  2ab  +  &2»  has 
already  been  subtracted  from  the  given  expression,  so  we 
Bhould  divide  the  remainder  by  twice  the  first  term,  i.e.,  by 
2(a  +  b),  for  a  new  term  of  the  root.  Then  for  a  new 
subtrahend  we  should  multiply  the  sura  of  2(«  +  b)  and  the 


EXAMPLES.  235 

new  term  by  the  new  term.     The  process  must  be  continued 
till  the  required  root  is  found. 

Hence  to  extract  the  square  root  of  a  polynomial,  we  have 
the  following 

Rule. 

Arrange  the  terms  according  to  the  powers  of  some  letter ; 
find  the  square  root  of  the  first  term  for  the  first  term  of  the 
square  root;  place  this  on  the  right,  and  subtract  its  square 
from  the  given  polynomial. 

Double  the  root  already  found  for  a  trial  divisor;  divide 
the  first  term  of  the  remainder  by  this  trial  divisor  for  the 
second  term  of  the  root,  and  annex  this  second  term  to  the  root 
and  cdso  to  the  trial  divisor  for  the  complete  divisor. 

Multiply  the  complete  divisor  by  the  second  term  of  the  root, 
and  subtract  the  product  from  the  remainder. 

If  there  are  other  terms  remaining,  repeat  the  process  until 
there  is  no  remainder,  or  until  all  the  terms  of  the  root  have 
been  obtained. 

EXAMPLES. 

1 .    Find  the  square  root  of 

4a;4  -  20a;3  +  37a:2  -  30a;  +  9  12a;2  -  bx  +  3. 


4a;4 


Ax'2  —  ox 


20a;3  +  37a;2 
20a;3  4-  25a;2 


4ar  -  10a;  +  3 


12a;2  -  30a;  +  9 
12a;2  -  30a;  +  9 


The  expression  is  arranged  according  to  the  descending 
powers  of  x. 

The  square  root  of  4a;4  is  2a;2,  and  this  is  placed  at  the 
right  of  the  given  expression  for  the  first  term  of  the  root. 
By  doubling  this  we  obtain  4a;2,  which  is  the  trial  divisor. 
The  second  term  of  the  root,  —5a;,  is  obtained  by  dividing 
—  20x3,  the  first  term  of  the  remainder,  by  4a;2,  and  this  new 
term  has  to  be  annexed  both  to  the  root  and  divisor.     Next 


236  EXAMPLES. 

multiply   the   complete   divisor   by   —  5x   and   subtract   the 
product  from  the  first  remainder. 

We  then  double  the  root  already  found  and  obtain 
4x2  —  lOx*  for  a  new  trial  divisor.  Dividing  12ar,  the  first 
term  of  the  remainder,  by  4a2,  the  first  term  of  the  divisor, 
we  get  3,  which  we  annex  both  to  the  root  and  divisor.  We 
now  multiply  the  complete  divisor  by  3  and  subtract.  There 
is  no  remainder,  and  the  root  is  found. 
2.    Find  the  square  root  of 

15a2x4  -  Gaz5  +  xG  -  20asx3  +  a6  +  lhah?  -  6a5x. 
Arrange  in  descending  powers  of  x. 
x6-6ax5+l5a2xi-20a3x3+loa4x2-6a5x+ae  |a;3-3ff.r2+3a2.r-a3 
x« 


2x3-Mx" 


— 6ax5+locrx* 
— Gax5+  9a2x4 


2xz-6ax2+3a?x 
2x8—  Qax2+6crx 


6a2a4-20a3a:3+15a4ar 
6a2x4-18a3.c8+  9«4ar 


-  2a3x-3+  6aV-6a5.?+a6 

-  2a3.r3+  6a4a:2-6a8a;+a6 


Note. —All  even  roots  admit  of  a  double  sign  (Art.  111).  Thus 
the  square  root  of  a2  +  2ab  +  &a  is  either  a  +  b  or  —  a  —  b,  as 
may  be  verified.  In  the  process  of  extracting  the  square  root  of 
a2  +  2ab  +  &2,  we  begin  by  extracting  the  square  root  of  a2,  and  this 
may  be  either  a  or  —a.  If  we  take  the  latter,  and  continue  the 
operation  by  the  ride  as  before,  we  shall  obtain  —  a  —  b.  A  similar 
remark  holds  in  every  other  case.  Thus,  in  Ex.  2  the  square  root  of 
the  first  term  x6  is  either  x8  or  —  xz.  If  we  take  the  latter,  and  continue 
the  operation  as  before,  we  shall  obtain  —  x3  +  3a.r2  —  Za*X  +  a*. 

Since  the  fourth  power  is  the  square  of  the  square,  the 
fourth  root  of  an  expression  may  be  found  by  extracting 
the  square  root  of  the  square  root.  Similarly  the  oiijhth  root 
may  be  found  by  three  successive  extractions  of  the  square 
root;  and  so  on. 

For  example,  required  the  fourth  root  of 

16a:4  -  9 Gar5//  +  216x-y2  -  2\(Sxif  +  M//'. 


EXAMPLES. 


237 


By  the  rule  we  find  that  the  square  root  is  Ax2  —  12a?/  4-  9.?/2 ; 
and  the  square  root  of  this  is  2x  —  %,  which  is  therefore 
the  fourth  root  of  the  given  expression. 

3.    Find  the  square  root  of 


24 


16?/2       8x      x*       32y 
x1         y        y2        x  ' 


Arranging  in  descending  powers  of  y,  we  have 


I6j/2 

,.2 


l  "//    32?/    ,    g^ 


8x   ,   x* 

y      y1 


-  4 


x2 


X 

-  4 

_32y+  24 

X 
X 

8 

c               y 

8  - 
8  - 

8x       x2 

y      f 

8x       x2 

y      y1 

Here  the  second  term  of  the  root,  —4,  is  found  by  the 
rule  as  usual,  i.e 

term,  -,  is  found  by  dividing  8  by  -^ 
y  x 


by  dividing ^  by  — ,  and  the  third 

xx 


EXAMPLES. 

Find  the  square  root  of  each  of  the  following 


4.  25a;2  -  30a?/  +  dif. 

5.  4a;4  -  12a3  +  29a;2  -  30a-  +  25. 

6.  1  -  10a  +  27a-2  -  10a3  +  a4. 

7.  4a2 +  9y2+  25z2+  12ay  —  SOyz  —  20az. 

8.  34a3-22a4+a6+ 121a2- 374a +  289. 


Ans.  5a  —  3?/. 
2a2  -  3a  +  5. 

1  -  5a  +  a2. 

2a  +  3y  —  5z. 
a3-lla-f  17. 


*  The  reason  for  this  arrangement  will  appear  in  Chap.  XIII. 


JLns. h  2. 

32/ 

rt4    .    x'2    ,    a3               .a2 
4        a2       05                   or 

-  2. 

a2       a       a; 
2         x       a 

238  SQUARE   ROOT   OF  ARITHMETIC  NUMBERS. 


10. 


113.  Square  Root  of  Arithmetic  Numbers. —  The 

rule  which  is  given  in  Arithmetic  for  extracting  the  square 
root  of  a  number  is  based  upon  the  method  explained  in  Art. 
112. 

Since  1  =  l2,  100  =  102,  10000  =  1002,  1000000  =  10002, 
and  so  on,  it  follows  that  the  square  root  of  a  number 
between  1  and  100  is  between  1  and  10;  the  square  root  of 
a  number  between  100  and  10000  is  between  10  and  100 ; 
the  square  root  of  a  number  between  10000  and  1000000 
is  between  100  and  1000,  and  so  on.  That  is,  the  square 
root  of  a  number  of  one  or  two  figures  consists  of  only  one 
figure  ;  the  square  root  of  a  number  of  three  or  four  figures 
consists  of  two  figures  ;  the  square  root  of  a  number  of  five 
or  six  figures  consists  of  three  figures ;  and  so  on.  Hence 
the 

Rule. 

If  a  point  is  placed  over  every  second  figure  in  any  number, 
beginning  with  the  units'  place,  the  number  of  points  toill  show 
the  number  of  figures  in  the  square  root. 

Find  the  square  root  of  5329. 

Point  the  number  according  to  the  rule.  Thus,  it  appears 
that  the  root  consists  of  two  places  of  figures,  i.e.,  of  tens 
and  units.     Let  a  denote  the 

value  of  the  figure  in  the  tens'  5321)  (70  +  3  =  73. 

place  of  the  root,  and  b  that  4900 

in  the  units'  place.  Then  a 
must  be  the  greatest  multiple 
of  10  whose  square  is  less  than 
5300;  this  we  find  to  be  70.  Subtract  a2,  i.e.,  the  square 
of  70,  from  the  given  number,  and  the  remainder  is  429, 
which  must  equal  (2a  +  b)b.     Divide  tins  remainder  by  the 


140  + 


429 
429 


SQUARE  ROOT   OF  ARITHMETIC  NUMBERS.  239 

trial  divisor,  2a,  i.e.,  by  140,  and  the  quotient  is  3,  which 
is  the  value  of  6.  Then  the  complete  divisor,  2a  +  6,  is 
140  +  3  =  143,  and  (2a  +  6)6,  that  is,  143  x  3  or  429 
is  the  number  to  be  subtracted ;  and  as  there  is  now  no 
remainder,  we  conclude  that  70  +  3  or  73  is  the  required 
square  root. 

In  squaring  the  tens,  and  also  in  doubling  them,  the 
ciphers  are  omitted  for  the  sake  of  brevity,  though  they  are 
understood.  Also  the  units'  figure  is  added  to  the  double  of 
the  tens  by  merely  writing  it  in  the  units'  place.  The  actual 
operation  is  usually  performed  as  follows  : 

If  the  root  consists  of  three  places  of  figures,  let  a  repre- 
sent the  hundreds  and  6  the  tens  ;  then  hav- 
ing obtained  a  and  6  as  before,  let  a  represent  5329(73 
the  hundreds  and  tens  as  a  new  value  ;  and  49 
find  a  new  value  of  6  for  the  units ;  and  in    143  \  429 
general,  let  a  represent  the  part  of  the  root  429 
already  found. 

Hence  for  the  extraction  of  the  square  root  of  a  number, 
we  have  the  following 

Rule. 

Separate  the  given  number  into  periods  of  two  figures  each, 
by  pointing  every  second  figure,  beginning  at  the  units'  place. 

Find  the  greatest  number  ivhose  square  is  contained  in  the 
left  period,  and  place  it  on  the  right;  this  is  the  first  figure 
of  the  root  ;  subtract  its  square  from  the  first  period,  and  to 
the  remainder  bring  down  the  next  period  for  a  dividend. 

Double  the  root  already  found  for  a  trial  divisor,  and  see 
how  many  times  it  is  contained  in  the  dividend,  omitting 
the  last  figure,  and  annex  the  result  to  the  root  and  cdso  to  the 
tried  divisor. 

Multiply  the  divisor  thus  completed  by  the  figure  of  the  root 
last  obtained,  and  subtract  the  product  from  the  dividend. 

If  there  are  more  periods  to  be  brought  down,  continue  the 
operation  as  before,  regarding  the  root  already  found  as  one 
term. 


240  SQUARE   ROOT   OF  A   DECIMAL. 

Extract  the  square  root  of  132496,  and  1024G401. 
1.    132496(364  2.    10246401(3201 

9  9 

66)424  62)124 

396  124 


724)2896  6401)6401 

2896  6401 

As  the  trial  divisor  is  an  incomplete  divisor,  it  is  sometimes 
found  that  the  product  of  the  complete  divisor  by  the  corre- 
sponding figure  of  the  root  exceeds  the  dividend.  In  such 
a  case  the  last  root  figure  must  be  diminished.  Thus,  in 
Ex.  1,  after  finding  the  first  figure  of  the  root,  we  are  re- 
quired by  the  rule  to  divide  42  by  6  for  the  next  figure  of 
the  root,  so  that  apparently  7  is  the  next  figure.  On  multi- 
plying however  67  by  7  we  obtain  a  product  which  is  greater 
than  the  dividend  424,  which  shows  that  7  is  too  large,  and 
we  accordingly  try  6,  which  is  found  to  be  correct. 

The  student  will  observe  in  Ex.  2  that,  in  consequence  of 
the  dividend,  exclusive  of  the  right  hand  figure,  not  contain- 
ing the  trial  divisor,  64,  we  place  a  cipher  in  the  root  and 
also  at  the  right  of  the  trial  divisor  64,  making  it  640 ;  we 
then  bring  down  the  next  period  and  proceed  as  before. 

114.  Square  Root  of  a  Decimal.  —  The  rule  for 
extracting  the  square  root  of  a  decimal  follows  from  the  rule 
of  Art.  113.  If  any  decimal  be  squared  there  will  be  an 
even  number  of  decimal  places  in  the  result;  thus  (.2o)2  = 
.0625,  and  (.111)-  =  .012321.  Therefore  there  cannot  be 
an  exact  square  root  of  any  decimal  which  has  an  odd 
number  of  decimal  places. 

The  square  root  of  32.49  is  one-tenth  of  the  square  root 
of  3249.  Also  the  square  root  of  .0361  is  one-hundredth  of 
that  of  361.  Hence,  for  the  extraction  of  the  square  root 
of  a  decimal,  we  have  the  following 


SQUARE  ROOT  OF  A   DECIMAL.  241 

Rule. 

Separate  the  given  number  into  periods  of  two  figures  each, 
by  putting  a  point  over  every  second  figure,  beginning  at  the 
units'  place  and  continuing  both  to  the  right  and  to  the  left  of 
it;  then  proceed  as  in  the  extraction  of  the  square  root  of  in- 
tegers, and  point  off  as  many  decimal  places  in  the  result  as 
there  are  periods  in  the  decimal  part  of  the  proposed  number. 

If  there  be  a  final  remainder  in  extracting  the  square  root 
of  an  integer,  it  indicates  that  the  given  number  has  not  an 
exact  square  root.  We  may  in  this  case  place  a  decimal 
point  at  the  end  of  the  given  number,  and  annex  any  even 
number  of  ciphers,  and  continue  the  operation  to  any  desired 
extent.  We  thus  obtain  a  decimal  part  to  be  added  to  the 
integral  part  already  found. 

Also,  if  a  decimal  number  has  no  exact  square  root,  we 
may  annex  ciphers  and  obtain  decimal  figures  in  the  root  to 
any  desired  extent. 

Find  the  square  root  of  12  ;  and  also  of  .4  to  three  deci- 
mal places. 


12.000000(3, 
9 

64)300 
256 

,464 

.400000(.632 
36 

123)400 
369 

686)4400 
4116 

1262)3100 
2524 

6924)28400 
27696 

Note.  —  We  see  here  in  what  sense  we  can  be  said  to  approximate 
Vo  the  square  root  of  a  number.  The  square  of  3.464  is  less  than  12, 
and  the  square  of  3.465  is  greater  than  12.  Also  the  square  of  .032  is 
less  than  .4,  and  the  square  of  .633  is  greater  than  .4. 

No  fraction  can  have  a  square  root  unless  the  numerator  and 
denominator  are  both  square  numbers  when  the  fraction  is  in  its  low- 
est terms.  But  we  may  approximate  to  the  square  root  of  a  fraction 
to  any  desired  extent.     Thus, 


242  SQUARE  ROOT   OF  A   DECIMAL. 

Let  it  be  required  to  find  the  square  root  of  -f . 

Here  (Art.  Ill)  i/|  =  zjL 

Therefore  we  find  the  square  root  of  5  and  also  of  7,  ap- 
proximately, aud  divide  the  former  by  the  latter. 

Or,  we  may  reduce  the  fraction  f  to  a  decimal  to  any 
required  degree  of  approximation,  and  obtain  the  square 
root  of  this  decimal. 

Otherwise  thus : 

~yTl       V^5  x  7       055  . 


vl-vf 


X   7       yj-t   x  7  7 

then  find  the  square  root  of  35  approximately,  and  divide 
the  result  by  7.  Either  of  these  last  methods  is  preferable 
to  the  first. 

If  the  square  root  of  a  number  consists  of  2n  +  1  figures,  when 
the  first  n  +  1  of  these  have  been  obtained  by  the  ordinary  method,  the 
remaining  n  may  be  obtained  by  division. 

Let  N  represent  the  given  number;  a  the  part  of  the  square  root 
already  found,  i.e.,  the  first  n  +  1  figures  found  by  the  rule,  with  n 
ciphers  annexed;  and  x  the  part  of  the  root  which  remains  to  be 
found. 

Then  y/jv  =  a  +  x; 

N  =  a'2  +  2ax  +  x2; 

.-.   £=*  =  .+  * (i) 

2a  2« 

Now  N  —  a-  is  the  remainder  after  n  +  1  figures  of  the  root,  rep- 
resented by  a,  have  been  found;  and  2a  is  the  corresponding  trial 
divisor.     We  see  from  (1)  that  N  —  a2  divided  by  2a  gives  x,  the  rest 

x2 
of  the  square  root  required,  increased  by  — . 

Now  —  is  a  proper  fraction,  so  that  by  neglecting  the  remainder 
2« 
arising  from  the.  division,  we  obtain  x,  the  rest  of  the  root.  For,  X 
contains  n  figures  by  supposition,  so  that  x2  cannot  contain  more  than 
_'//  figures ;  but  a  contains  2n  +  1  figures  (the  last  n  of  which  are  ciphers) 
and  thus  2a  contains  2n  +  I  figures  at  least;  therefore  ^-  is  a  proper 

fraction. 

Prom  ibis  investigation,  by  putting  n  =  1,  we  Bee  thai  al  leasl  two 
of  the  figures  of  ;i  square  root  must  have  been  obtained  in  order  that 


CUBE  ROOT   OF  A   POLYNOMIAL.  243 

the  method  of  division  may  give  the  next  figure  of  the  square  root 
correctly. 

We  will  apply  this  method  to  finding  the  square  root  of  290  to  five 
places  of  decimals.  We  must  obtain  the  first  four  figures  in  the  square 
root  by  the  ordinary  method ;  and  then  the  remaining  three  may  be 
found  by  division. 

290  (17.02 
1 
27)  190 


3402)  10000 
6804 
3196 
We  now  divide  the  remainder  3196,  which  is  N  —  a2,  by  twice  the 
root  already  found,  3404,  which  is  2a,  and  obtain  the  next  three 
figures.     Thus, 

3404)  31960  (938 


13240 
10212 


30280 

27232 

3048 

Therefore  to  five  places  of  decimals,  y/090  =  17.02938. 

In  extracting  the  square  root,  the  student  will  observe  that  each 
remainder  brought  down  is  the  given  expression  minus  the  square  of 
the  root  already  found,  and  is  therefore  in  the  form  2V  —  a2, 

EXAMPLES. 

Find  the  square  roots  of  the  following  numbers : 

5.  .835396.    Am.   .914. 

6.  1522756.       1234. 

7.  29376400.       5420. 

8.  384524.01.     620.1. 

115.  Cube  Root  of  a  Polynomial.  —  Since  the  cube 
of  a  +  b  is  «3  -f  Serb  +  oab2  +  ft8,  the  cube  root  of 
a3  +  Serb  +  Sab'2  4-  b3  is  a  +  b.  We  may  deduce  a 
general  rule  for  the  extraction  of  the  cube  root  of  a  poly- 


1. 

15129. 

Ans.  123. 

2. 

103041. 

321. 

3. 

3080.25. 

55.5. 

4. 

41.2164. 

G.42. 

244  CUBE   ROOT   OF  A   POLYNOMIAL. 

nomial  by  observing  in  what  manner  a  -f-  b  may  be  derived 
from  a8  +  Serb  +  Sab2  +  b3. 

Arrange  the  terms  of  the  cube  according  to  the  descending- 
powers  of  a\  then  the  first  term  is  a3,  and  its  cube  root  is  a, 
which  is  the  first  term  of  the  required  root.  Subtract  its 
cube,  a8,  from  the  given  expression,  and  bring  down  the 
remainder  3a2&  +  Sab'2  -+-  b3.  Thus,  o,  the  second  term  of 
the  root,  will  be  the  quotient  when  3a2o,  the  first  term  of  the 
remainder,  is  divided  by  3a'2,  i.e.,  by  three  times  the  square 
of  the  first  term  of  the  root. 

Also,  since  Sa2b  +  3a&2  +  b3  =  (3a2  +  3a?>  +  &2)o,  we 
add  to  the  trial  divisor  Sab  -f  &2,  i.e.,  three  times  the 
product  of  the  first  term  of  the  root  by  the  second,  plus 
the  square  of  the  second,  and  we  have  the  complete  divisor 
3a2  +  Sab  -f-  b'2 ;  multiply  this  complete  divisor  by  6,  and 
subtract  the  product,  Sarb  -j-  Sab'2  +  b3,  from  the  remainder, 
and  the  operation  is  completed. 

The  work  may  be  arranged  as  follows  : 

a3  +  3a26  +  oab2  +  b3\a  +  b 
a3 


3a2  +  Sab  +  V2 


S,rb  +  Sab2  +  b3 
:\a2b  +  Sab2  +  b3 


If  there  were  more  terms,  we  should  proceed  with  a  +  b 
as  we  have  done  with  a ;  its  cube,  a3  +  :)<rb  +  dab3  +  b3, 
has  already  been  subtracted  from  the  given  .expression,  so 
we  should  divide  the  remainder  by  three  times  the  square  of 
the  first  term,  i.e.,  by  3(a  +  6)'J,  for  a  new  term  of  the  root, 
c  say.  Then  for  a  new  complete  divisor  we  would  have 
3(a  +  b)2  +  3(a  +  b)c  +  cs;  and  this  multiplied  by  c 
would  give  us  a  new  subtrahend  ;  and  so  ou.  Hence  the 
following 

Rule. 

Arrange  the  terms  according  to  the  poioers  of  some  letter; 
find  the  cube  root  of  ihc  first  term  for  the  first  term  of  the 
cube  root;  and  subtract  its  cube  from  the  given  polynomial. 


EXAMPLES. 


245 


Take  three  times  the  square  of  the  root  already  found  for  a 
trial  divisor;  divide  the  first  term  of  the  remainder  by  this 
trial  divisor  for  the  second  term  of  the  root;  annex  this  second 
term  to  the  root,  and  complete  the  divisor  by  adding  to  the  trial 
divisor  three  times  the  product  of  the  first  and  second  terms 
of  the  root  and  the  square  of  the  second  term. 

Multiply  the  complete  divisor  by  the  second  term  of  the  root, 
and  subtract  the  product  from  the  remainder. 

If  there  are  other  terms  remaining,  take  three  times  the 
square  of  the  part  of  the  root  already  found  for  a  new  trio* 
divisor;  and  continue  the  operation  until  there  is  no  remain- 
der, or  until  all  the  terms  of  the  root  have  been  obtained. 


EXAMPLES. 

1.    Find  the  cube  root  of  8a;3  -  3Gar?/  +  54a,*?/2  -  27?/8. 
The  work  may  be  arranged  as  follows  : 

8a-3-36a-'2y/+54a-?/2-27?/3 1 2a;  -  3y 

8a;3 


3  (2a;)2  =12ar 

8(2a-)(~8i0  =        -ISxy 

(Syy=  +dy2 


12x2-18xy+9y' 
2.    Find  the  cube  root  of 


-  36x-y + bAxy1— 2  7y3 
-36x2y+o4xy*-27ys 


I3.j-4a._2ar1 


27+108B+   90a;2-   80b3-60b4+48b5-8b6 

27 


27  +  3Ga;  +  16a-2 


108b+   90a;2-   80b8 
108b+144b2+   64a;3 


27  +  72b+48b2 

-18a;2 -24a;3 


+  4b4 


27  +  72b+30b2-24b3+4b4 


54b2— 144b3— G0b4+48b5— 8b6 


.54a2— 144a8- 600*+ 48a5— 8a6 


Explanation.  —  The  root  is  placed  above  the  given  expression  for 
convenience.     When  we  have  obtained  two  terms  in  the  root,  3  +  4x, 


246  CUBE  ROOT  OF  ARITHMETIC  NUMBERS. 

we  form  the  second  divisor  as  follows:  take  3  times  the  square  of  the 
root  already  found  for  the  trial  divisor,  27  +  72x  +  48x2;  divide  —54.x'2, 
the  first  term  of  the  remainder,  by  27,  the  first  term  of  the  trial 
divisor;  this  gives  the  third  term  of  the  root,  — 2x2.  To  complete  the 
divisor  we  add  to  the  trial  divisor  3  times  the  product  of  (3  +  4x)  and 
— 2x2,  and  also  the  square  of  — 2x2.  Now  multiply  the  complete  divisor 
by  — 2x2  and  subtract;  there  is  no  remainder  and  the  root  is  found. 

Find  the  cube  root  of  each  of  the  following  : 

3.  a3  +  3a2  +  3a  +  1.  Ans.  a  +  1. 

4.  G4a3  -  144a26  +  108a&2  -  27b3.  4a  -  3b. 

5.  xG  +  3a5  +  Gx4  +  7x3  +  6x2  +  3x  +  1.  x2  +  x  +  1. 

6.  l-6x-+21a;2-44ar5+G3:z4-54a;5+27a:6.  l-2x+3x~. 

116.  Cube  Root  of  Arithmetic  Numbers. — The 

rule  which  is  given  in  Arithmetic  for  extracting  the  cube  root 
of  a  number  is  based  upon  the  method  explained  in  Art.  115. 

Since  1  =  l3,  1000  =  103,  1000000  =  1003,  and  so  on,  it 
follows  that  the  cube  root  of  a  number  between  1  and  1000 
is  between  1  and  10;  the  cube  root  of  a  number  between 
1000  and  1000000  is  between  10  and  100  ;  and  so  on.  That 
is,  the  cube  root  of  a  number  of  one  or  two  or  three  figures 
consists  of  only  one  figure  ;  the  cube  root  of  a  number  of 
four  or  Jive  or  six  figures  consists  of  two  figures  ;  and  so  on. 
Hence  the 

Rule.  If  a  point  is  placed  over  every  third  figure  in  any 
number,  beginning  with  the  imits'  place,  the  number  of  points 
ivill  show  the  number  of  figures  in  the  cube  root. 

Find  the  cube  root  of  G14125. 

Point  the  number  according  to  the  rule.  Thus  it  appears 
that  the  root  consists  of  two  places  of  figures,  i.e.,  of  tens 
and  units.  Let  a  denote  the  value  of  the  figure  in  the  tens' 
place  of  the  root,  and  b  that  in  the  units'  place.  Then  a 
must  be  the  greatest  multiple  of  10  whose  cube  is  less  than 
G14000  ;  this  we  find  to  be  80.  Subtract  a3,  i.e.,  the  cube 
of  80,  from  the  given  number,  and  the  remainder  is  102125, 
which  must  equal  (3a2  +  Bab  +  b2)b.  Divide  this  remainder 
by  the  trial  divisor,  3a2,  i.e.,  by  10200,  and  the  quotient  is  .">, 


CUBE  ROOT  OF  ARITHMETIC  NUMBERS.  247 

which  is  the  value  of  b.  Then  adding  Sab,  or  1200,  and  62, 
or  25,  to  the  trial  divisor  3a2,  or  19200,  we  obtain  the  com- 
plete divisor  20425  ;  and  multiplying  the  complete  divisor 
by  5  and  subtracting  the  product  102125,  there  is  no  remain- 
der.    Therefore  85  is  the  required  cube  root. 

614125  1 80  +  5 
512000  ~~ 


3a2  =  3(80)2       =  19200 

3ab  =  3(80)  (5)  =     1200 

b*  =  (5)2  =         25 


20425 


102125 


102125 


In  cubing  the  tens  the  ciphers  are  omitted  for  the  sake  of 
brevity,  though  they  are  understood. 

If  the  root  consists  of  three  places  of  figures,  let  a  repre- 
sent the  hundreds  and  b  the  tens,  and  proceed  as  before. 
See  Art.  113. 

Hence  for  the  extraction  of  the  cube  root  of  a  number, 
we  have  the  following 

Rule. 

Separate  the  given  number  into  periods  of  three  figures 
each  by  pointing  every  third  figure,  beginning  at  the  units' 
place. 

Find  the  greatest  number  whose  cube  is  contained  in  the  left 
period,  and  place  it  on  the  right;  this  is  the  first  figure  of  the 
root;  subtract  its  cube  from  the  first  period,  and  to  the  re- 
mainder bring  down  the  next  period  for  a  dividend. 

Take  three  times  the  square  of  the  root  already  found  for 
a  trial  divisor,  and  see  hoiv  many  times  it  is  contained  in  the 
dividend,  omitting  the  last  tivo  figures,  and  annex  the  residt 
to  the  root.  Add  together,  the  trial  divisor  ivith  two  ciphers 
annexed;  three  times  the  product  of  the  last  figure  of  the  root 
by  the  rest,  with  one  cipher  annexed;  and  the  square  of  the 
last  figure  of  the  root. 


248  CUBE  ROOT   OF  A    DECIMAL.       ■ 

Multiply  the  divisor  thus  completed  by  the  figure  of  the  root 
last  obtained,  and  subtract  the  product  from  the  dividend. 

If  there  are  more  periods  to  be  brought  cloivn,  the  operation 
must  be  repeated,  regarding  the  root  already  found  as  one  term. 

Extract  the  cube  root  of  109215352. 

109215352(478 

64 


3(4)2 

= 

4800 

3(4) (7) 

= 

840 

(<y 

= 

49 

5(589 


15215 


39823 


3(47)2  =  G62700 

3(47) (8)    =     11280 

(8)2  =  G4 


674044 


5392352 


5392352 


After  finding  the  first  figure  of  the  root  we  fire  required  by 
the  rule  to  divide  452  by  48  for  the  next  figure  of  the  root, 
so  that  apparently  8  or  9  is  the  next  figure.  On  trial  we  find 
that  these  numbers  are  too  large  ;  so  we  try  7,  which  is  found 
to  be  correct.  As  in  the  case  of  the  square  root  (Art.  113), 
we  are  liable  occasionally  to  try  too  large  a  figure,  especially 
at  the  early  stages  of  the  operation. 

116a.  Cube  Root  of  a  Decimal.  —  If  the  cube  root 
have  any  number  of  decimal  places,  the  cube  will  have  three 
times  as  many.  Hence  for  the  extraction  of  the  cube  root 
of  a  decimal,  we  have  the  following 

Rule. 

/Separate  the  given  number  into  periods  of  three  figures 
each,  by  putting  a  point  over  every  third  figure,  beginning  at 
the  units'  place  and  continuing  both  to  the  right  and  to  the  left 
of  it;  then  proceed  as  in  the  extraction  of  the  cube  root  of 
integers,  and  point  off  as  many  decimal  places  in  the  result 
as  there  are  periods  in  the  decimal  part  of  the  proposed 
number. 


CUBE   ROOT   OF  A   DECIMAL. 


249 


If  there  be  a  final  remainder  in  extracting  the  cube  root 
of  any  number,  integral  or  decimal,  it  indicates  that  the 
number  has  no  exact  cube  root.  We  may  in  this  case,  as  in 
the  extraction  of  the  square  root  (Art.  114),  annex  any 
number  of  ciphers,  and  continue  the  operation  to  any  desired 
extent. 

Extract  the  cube  root  of  1481.544 


300 

481 

30 

1 

331 

331 

3G300 

150544 

1320 

16 
37636 

150544 

The  cube  root  is  11.4. 

If  the  cube  root  of  a  number  consists  of  2n  +  2  figures,  when  the 
first  n  +  2  of  these  have  been  obtained  by  the  ordinary  method, 
the  remaining  n  may  be  obtained  by  division. 

Let  N  denote  the  number;  a  the  part  of  the  cube  root  already 
found,  i.e.,  the  first  n  +  2  figures  found  by  the  rule,  with  n  ciphers 
annexed;  x  the  remaining  part  of  the  root. 
Then  _ 

■fr N  —  a  A-  x; 
.:    N  =  a3  +  3a2x  +  3ax~  +  a;3; 


X 


=  x  + 


;n 


3a2  '  a    "  3a2 

Now  N  —  a3  is  the  remainder  after  n  +  2  figures  of  the  root, 
represented  by  a,  have  been  found;  and  3a'2  is  the  corresponding  trial 
divisor.    We  see  from  (1)  that  iV  —  a3  divided  by  3a2  gives  x,  the  rest 


of  the  cube  root  required,  increased  by \- 


:;«- 


This  latter  expres- 


sion is  a  proper  fraction,  so  that  by  neglecting  the  remainder  arising 
from  the  division,  we  obtain  x,  the  rest  of  the  root  required.  This 
may  be  shown  as  follows : 


250  EXAMPLES. 

By  supposition,  x  contains  n  figures,  and  a,  2n  +  2  figures. 
.*.     a-  <  10",  and         a>  10*»+1: 


x2 
a 

.     102» 

< 

Tu- 

Also 

Xs 

3a2 

IO3" 

or  < 

1 

<3  x  104n  +  2 

3.x 

10" 

+  U 

X2 

a 

X8 

+  3^<^  + 

1 

-a' 

3  x 

10^ 

and  is 

therefore 

a  proper  fraction. 

EXAMPLES. 

Show  that 

1.  (-2a5)4  =  16a20. 

2.  (-a4)5  =  -a20. 

3.  (-3aW)3  =  -27<rV5c3. 

4.  (— 5a'W)3  =  -125a°6V2. 

5.  (-3a762c)4  =  81a2W. 

Find  the  value  of 

G.    (-ax*  +  6?/3)2.  Ans.  aV  -  2a6x4?/8  +  by. 

7.  (2a4  +  3&3)2.  4a8  +  12a463  +  9b6. 

8.  (2a2  -  362)3.  8a«  -  3Ga4^2  +  54a2i4  -  2766. 

9.  (2x  +  3)4.  lGx4  +  %x3  +  216x2  +  216a;  +  81. 

10.  (1  +  x)5  -  (1  -  a-)5-  2(5aj  +  10x3  +  aj6). 

11.  (1  —  3«  +  3a/2)2.  1  -  6a;  +  15a?2  -  18x3  +  9a^. 

12.  (2  +  3x+4x2)2+(2-3x  +  4x2)2.     2(4  +  25x2+lGx4). 

13.  (1  +  3a;  +  2a?)*. 

Ans.  1  +  9a;  +  33a;2  +  63a?  +  GGx4  +  3Gx5  +  8aj". 

14.  (2  +  3x  +  4x2)3  -  (2  -  3x  +   ix'-')3. 

J/i.s.  2(3Gx  +  171x3  +  lll.r'). 

15.  (1  +  4x  +  6aJa  +  <lx3  +  x4)2. 

Ans.  l+.Sx-r-2,Sx'J  +  r)i;./:)+70.r1  +  5Gx5  +  28x6+8x7+x8. 


EXAMPLES. 


251 


Show  that 


16.   V32zy°  =  2xf. 


17.    V256a8o;64  =  2ax\ 


IS.    y~32xwy15  =  -2xY 
19.    1°/^ 


a" 


Find  the  square  roots  of  the  following  expressions : 

20.  9xA  -  12a;3  -  2a;2  -f  4a;  -f  1.       .4ns.  3a;2  —  2a;  —  1. 

21.  16a;6  +  16a;7  -  4a;8  -  4a;9  +  x10.         4a;3  +  2a;4  -  a;5. 

22.  25a;4-30aa;3+49a2a;2-24a3a;+16a4.     5a;2 -  3ax  -f  4a2. 

23.  a;4  -  4a:3  +  8a:  +  4.  ar  -  2a;  -  2. 

24.  a;6+  4ao;5  -  10a3o;3  +  4a5x  +  a6,    a;3  +  2aa;2  -  2a2a;  -  a3. 

25.  a;4  -  2aa;3  +  (a2  +  262)a;2  -  2a62a;  +  b4.    x1  -ax  +  b\ 

26.  16  -  96a;  +  216a;2  -  216a;3  +  81a;4.     4  -  12a;  +  9a;2. 

27.  9o;6-12a;5+22a;4  +  a;2  +  12a;  +  4.     3a;3  -  2a;2  +  3a;  +  2. 

28.  4a;8  -  Ax6  -  7a;4  +  4a;2  +  4.  2a;4  -  a;2  -  2. 

29.  1  —  xy  —  \5-xhf  +  2x3y3  +  4a;4?/4.         1  -  \xy  -  2x2y2. 


30.    ^!  +  4a;2  +  —  +  — 
4  3         9 


Find  the  square  roots  of 

81.  165649.         Ans.  407. 

32.  384524.01.  621.1. 

33.  4981.5364.  70.58. 

34.  .24373969.  .4937. 

35.  144168049.        12007. 


2a:3 


4aa; 
3  ' 


2a;  + 


36.  .5687573056.  Ans.  .75416. 

37.  3.25513764.  1.8042. 

38.  4.54499761.  2.1319. 

39.  196540602241.       443329. 


Find  the  square  roots  of  the  following  to  five  decimals : 


40.  .9.  Ans.  .94868. 

41.  6.21.  2.49198. 

42.  .43.  .65574. 


43.  .00852. 

44.  17. 

45.  129. 


Ans.  .09233. 

4.12310. 

11.35781. 


252 


EXAMPLES. 


Find  the  cube  roots  of  the  following  expressions  : 

4G.    1 728a6 + 1728a4?/3  +  576aY+64?/9.     Ans.  12x2  +  Ay\ 

47.  x6  —  Sax5  +  5a3x3  —  3a6aj  —  a6.  a2  —  ax  —  a2. 

48.  8a6+48ca5  +  60c2x4  -  S0c3x3  -  90cV  +  108c5a  -  27r6. 

Ans.  2a2  +  4cx  —  3c'-. 

49.  1  -  9a  +  39a2  -  99a3  -f  156a4  -  144a5  +  64.x6. 

Ans.  1  —  3x  +  4.t2. 

50.  27a6  -  27a5  -  18a4  +  17a3  +  Gx2  -  3a  -  1. 

Ans.  3a2  —  a  —  1. 


51. 

x*  +  §*!  +  *^  _  t  +  §2?  _  2^  _  4. 

?/3        t/2         ?/        a3        a2         a 

^  +  2-^ 
2/               * 

52. 

65       6a       „       a3       3a2       3&2       63 
a  +   b                 bs        b*         a2        a3' 

|  - 1  +  5 

6                a 

Find  the  fourth  roots  of 

53.  1  +  4a  +  6a2  +  4a3  +  a4. 

54.  l-4a+  10a2-  16a3 -f-  19a4 


Ans.  1  +  a. 

16a5  +  10a6  -  4a7 + a8. 

Ans.  1  —  a  +  a2. 


Find  the  sixth  root  of 

55.  1  +  12a  +  60a2  +  160a3  +  240a4  +  192a6  +  64a6. 

Ans.  1  +  2a. 
Find  the  cube  roots  of 

56.  2628072.      Ans.  138. 

57.  3241792.  148. 

58.  60236.288.  39.2. 

59.  191.102976.         5.76. 


60.  .220348864.  Ans.  .604. 

61.  1371330631.  1111. 

62.  20910518875.        2755. 


THE   THEORY   OF  EXPONENTS  —  SURDS.  253 


CHAPTER    XIII. 

THE    THEORY     OF     E  X  PO  NE  NT  S  —  S  U  RD  S. 

117.  Exponents    that    are    Positive    Integers.  — 

Hitherto  we  have  supposed  that  an  exponent  was  always  a 
jjositive  integer.      Thus,  in  Art.  12,  we  defined  am  as  the 
product  of  m  factors  each  equal  to  a,  which  would  have  no 
meaning  unless  the  exponent  was  a  positive  integer. 
When  m  and  n  are  positive  integers,  we  have 
am=  a  ■  a  •  a  .  .  .  torn  factors  ; 
and        an=  « •  a  •  a  .  .  .  to  n  factors. 

.  • .  a"1  x  a" = (a  •  a  •  «...  to  m  factors)  x  (a  •  a  •  a . .  ton  factors) 
=  a-  a  •  a  .  .  ,  to m-\-n factors 
=  am+n  by  definition  (Art.  12) (1) 

Also  a™  +  *  =  £=  a  •  a  •  «  .  .  .  to  m  factors 
a"        a  •  a  •  a  ...  to  w  factors 

=  a  -  a  •  a  ...  to  m  —  ?i  factors 

=  «'"- (2) 

From  Art.  108  we  have 

KT  =  a"", (3) 

and  a"1  x  6m  =  («^)m (4) 

These  four  fundamental  laws  of  combining  exponents  are 
proved  directly  from  a  definition  which  has  meaning  only 
when  the  exponents  are  positive  and  integral. 

118.  Fractional  Exponents.  —  It  is  often  found  con- 
venient to  use  fractional  and  negative  exponents,  such  as  a^, 
a~5,  which  at  present  have  no  intelligible  meaning,  because 
we  cannot  write  a  1|  times  or  —5  times  as  a  factor.  It  is 
very  important  that  Algebraic  symbols  should  always  obey 


254  FRACTIONAL   EXPONENTS. 

the  same  laws  ;    and  to  secure  this  result  in  the  case  of 
exponents,  the  definition  should  be  extended  so  as  to  include 
fractional  and  negative  values.     Now  it  is  found  convenient 
to  give  such  definitions  to  fractional  and  negative  exponents 
as  will  make  the  relation 

am  x  an  =  am  +  n  ......   (1) 

ahvays  true,  whatever  m  and  n  may  be. 

To  find  the  meaning  of  a'2. 

Since  (1)  is  to  be  true  for  all  values  of  m  and  n,  we  must 
have 

ah  x  ah  =  a?  +  ?  =  a1  =  a. 

Thus  a^  must  be  such  a  number  that  its  square  is  a.  But 
the  square  root  of  a  is  such  a  number  (Art.  13).     Therefore 

ah  =  \Ja (2) 

To  find  the  meaning  of  ah. 

By  (1)      ah  x  ah  X  ah  =  aHi+i  =  al  =  a. 

Hence  ah  must  be  such  a  number  that  when  taken  three 
times  as  a  factor  it  produces  a  ;  that  is,  as  must  be  equivalent 
to  the  cube  root  of  a. 

.-.     «i  =  % (3) 

To  find  the  meaning  of  ah. 

By  (1)  ah  x  at  x  at  x  ah  =  a3. 

.-.     at  =  Va»     .     .     .     .   (4) 
To  find  the  meaning  of  an,  where  n  is  any  positive  integer. 

By(i) 

</ »  x  a »  x  a » X  .  .  .  to  n  factors  =  ah +  '<  + ■ +  •  •  ■ ton ,cr'"9  =  a1  =  a  ; 
therefore  a"  must  be  such  that  its  nth  power  is  a. 

a«  =  Vrt (5) 

m 

To  /mcZ  </ie  meaning  of  a" ,  where  m  and  n  are  any  positive 
integers. 


NEGATIVE  EXPONENTS.  253 

By  (1)      an  x  a"  x  an  x to  n  factors 

_+-+-+ to  n  terms 

_   an       n       n  _    am  . 

therefore  a"  must  be  equal  to  the  nth  root  of  am  ;  that  is, 

a"  =  V«m (6) 

i  i  x  ™ 

Also,  a*  x  a"  x  a"  X  ......  to  ??i  factors  =  a"  ; 

therefore  an  means  also  the  ??ith  power  of  a" ;    that  is,  from 
(5) 

m 

a-  =  (V«)m (7) 

Therefore  from  (6)  and  (7),  a"  =  V«m  =  (Va)m  •     .     .   (8) 

Hence  an  means  the  n"'  root  of  the  mth  power  of  a,  or  the 
mth  power  of  the  n"1  root  of  a;  that  is,  in  a  fractional 
exponent  the  numerator  denotes  a  power  and  the  denominator 
2  root. 

Examples.  x%  =  {/?,  a§  =  sfa1,  4*  =  ^4»  =  034  =  8. 

119.  Negative  Exponents.  —  ( 1 )  To  find  the  meaning 
of  o°. 

By  (1)  of  Art.  118,  a0  x  a"  =  a0  +  n  =  an; 

.-.  a°  =  a*  -s-  an  =  1  .  .  .  (1) 
Hence,  an?/  number  whose  exponent  is  zero  is  equal  to  1. 
(See  Art.  45). 

(2)  To  find  the  meaning  of  a-",  where  n  is  any  positive 
number. 

By  (1)  of  Art.  118,  a"  X  a~n  =  an~n  =  a0  =  1  [from  (1)]. 

Hence  a"  =  ,     and     a~n  =  —  .     .     .     .    (2) 

a-"  an 

Thus  we  see  that  any  quantity  may  be  changed  from  the 
numerator  to  the  denominator,  or  from  the  denominator  to 
the  numerator,  of  a  fraction,  if  the  sign  of  its  exponent  be 
changed. 


256 

Examples.  x~2  = 
a2b3 


EXAMPLES 

1         1 


x- 


l        k/~        1  2        s/~5 

=  as  =  \x ; =  x>  =  \x- ; 


.r;r 


a2bsx~hj- 


a  ~  ~b  ~  3xy 


27-1  =JL=      1 


8T-  =  i  =  i(by(8)°fArt-118)- 

27*       V272       V36       3" 


Otherwise  thus 


27*       (V27)2       32 


(3)    To  prove  that  a"1  -~  a"  =  «'"_n  /or  a?Z  values  of  m 
and  n. 


am  -7-  an  =  —  =  a'"  X  a-"  =  am~",  by  the  fundamental 


law. 


Examples,  a3  -=-  a5  =  a3-5  =  a    - 

a  -=-  a- J  =  a1+5  =  a"3' . 
2a*  X  a*  x6a_5 


9a~§  X  a* 

V^3  x  vV2 


|ai+l-5+f-i  _  |( 


3a 


a*  x  ^ 


Vy~-  X  Vx6       y    •'*  x  a$ 


x$    $yi+-s  =z  x°y  =  y. 


Note.  — It  appears  that  it  is  not  absolutely  necessary  to  introduce 
fractional  and  negative  exponents  into  Algebra,  since  they  merely 
supply  us  with  a  new  notation  in  addition  to  one  we  already  had.  It 
is  simply  a  convenient  notation,  which  the  student  will  learn  to 
appreciate  as  he  proceeds. 


EXAMPLES. 
Express  with  positive  exponents  : 


i      3  z 

1.    2x    hi    ■'.  Ans.  ~t~^"- 


2x*  X  3.):-1 

V*8 


3.    ^X^X^ina>gy 

4.  vsr^  +  VS.  -, 


TO  PROVE  THAT  (am)n  =  amn  IS  UNIVERSALLY  TRUE.    2o7 
Express  with  radical  signs  : 


5.    bcftx    $b    %. 


6.   a-*  x  2a' 


A, 


Wa_ 

\J.vb* 

_2_ 

Va"5" 


7.    a;-'  -r-  2a" 


8.    7a~8  x  3a-1. 


Ans. 


21 


120.  To  Prove  that  (a»)"  =  a""'  is  Universally  True 
for  All  Values  of  m  and  n. 

1.  iei  n  be  a  positive  integer,  and  m  have  any  value. 
Then  from  the  definition  of  a  positive  integral  exponent 

(a'")"  =  a"'  x  a"1  x  am  x to  n  factors 

qJU  +  m  +vi  + to  n  terms 

=  amn (1) 

2.  Let  n  be  a  positive  fraction   *-,  where  p  and  q  are 

positive  integers,  and  m  unrestricted  as  before.     Then 

(a'")n  =  (amy  =  V  («"')"  (Art.  118) 
=  VaTp  by  (1) 

=  aJ  (Art.  118) 

=  «"■" (2) 

3.  Let  n  be  negative,  and  equal  to  —p,  where  p  is  a  positive 
integer,  and  m  unrestricted  as  before.     Then 

(«'»)"  =  («»)-i>  =  7-^(Art'  119) 
=  —  by  (1)  and  (2) 

amp      J    V    t  V    * 

=  a-™*  =  «•»» (3) 

Hence  (am)n  =  (a")m  =  a""1  for  all  values  of  m  and  n. 

4.  Let  n  =  -. 
Then  we  have 


(am)"  =  (a")m  =  a" 


(4) 


258     TO  PROVE  THAT  ((lb)"  =  a"b"  FOR  ANY  VALUE  OF  n. 

That  is,  the  n'h  root  of  the  m'k  power  of  a  is  equal  to  the  mth 
power  of  the  nth  root  of  a. 

5.    Let  m  =  —  and  n  =  -. 
m  n 

11  11  _L_ 

(a"')n  =  (a")™  =  amn (5) 

That  is,  the  n'h  root  of  the  m'h  root,  or  the  m'h  root  of  the  ntA 
root  of  a  is  equal  to  the  mnth  root  of  a. 

Examples.   (b*)$  =  fi3x?  =  &'. 

[>-2)3]-3  =  (x-6)-5  =  a-6(-j)  =  x\ 

(fry  =  (3*)3  =  3*  =  ^3. 

Vv/27^3  =  [(27a;3)*]*  =  [(27a8) -3]*  =  ^3x. 

121.  To  Prove  that  (ab)"  =  a"b"  for  Any  Value  <U 

n.  —  This  has  already  been  shown  to  be  true  when  n  is  ft 
positive  integer  (Art.  108). 

1.  Let  n  be  a  p>ositive  fraction  £_,   lohere  p  and  q  a/e 

Q 
positive  integers.     Then 

(ab)n  =  (a&)«. 

Now  [(«&)*]'  =  («&)'  (Art.  120) 

=  apb"  (Art.  108) 
p  p 

=  (a«&«)«. 

P  21  1! 

.-.     (ab)q  =  aW () 

2.  Let  n  have  any  negative  value,  say  —r,  where  i   >s  a 
positive  integer.     Then 

(ab)"  =  (ab)-r  =  — !— 
K     >         K  (aby 

-JL-o-&-%.     ...        (2) 

a'b' 

which  proves  the  proposition  generally. 


EXAMPLES. 


259 


Iu  this  proof  the  quantities  a  and  b  are  wholly  unrestricted^ 
and  may  themselves  have  exponents. 

P  _  1 
Let  —  —  -.     Then  from  ( 1 )  we  have 

q       n  v  ' 

i  11 

(ab)n  =  a*b\ 


r.     \ab  =  \a  •  nyjb 


(3) 


That  is,  the  nth  root  of  the  product  is  equal  to  the  product 
of  the  n'h  roots. 

EXAMPLES. 

1.    («%-*)*  -^  {xhj-1)-*  =  x$y~%  -f-  x~^y->  =  xhj-1. 
Express  with  positive  exponents 


Ans, 


8.  yab-h-*  x  (a-^-^j-*) -*. 

9.  Va^  X  (a%-a)"ft. 


1 

4 
9a2a;2' 

16ac4. 


5.    (aty-»)«(»V)- 


Ans. 


6.  y/x^x-K 


7.    (4a 


9a2)- 


1 

.2a  +  36* 

xK 

3a.r 
2  ' 


:,;'' 


10.    (a~*Va;)-8  x  ^x-*Jar\ 


11.    ^(a  +  6)5  X  (a  +  &)-t. 


a  +  b. 


Rem.  —  Since  the  laws  of  the  exponents  *  just  proved  are  universally 
true,  all  the  ordinary  operations  of  multiplication,  division,  involution, 
and  evolution  are  applicable  to  any  expressions  which  contain  frac- 
tional and  negative  exponents. 


(Jailed  the  index  Itnvs. 


2G0  EXAMPLES. 

The  reason  for  the  arrangement  in  Ex.  3,  Art.  112,  may 
now  be  seen.     Thus  the  descending  powers  of  x  are 
.3     .2  ,111 

X  ■,  X  .   X.    1,    ,  — -,   — , 


as  may  be  seen  (Art.  119)  by  writing  the  terms  as  follows : 


12.    Multiply  3x-?  +  x  +  2xi  by  xs  —  2. 
Arrange  in  descending  powers  of  x. 

x    -f  2xi  +  3a;  ~* 

a;*  -  2 


xs  +  2x    +  3 

—  2x    —  4xt  —  6a;-* 


*t  -  4o;t  +  3      -  6a;-*. 

13.  Divide  3a;*2r*+B*—  3a?*2T*— 2T*  by  aj*+2T*— 2a;*#-*. 

Arrange  in  descending  powers  of  x. 
x\  —  2xhj~l  +  y~*[  a;*  —  3o5%-*  +  3a;*?/-*  —  y~^  \  a*  —  y~s 
»i  —  2aj%~6  +    xhj-'t 

—  x$y-$  +  2#*y~*  —  y* 

—  aj%-i  +  2x$y~i  —  y* 
Multiply 

14.  x§  +  y\  by  a$  —  ?/*.  j4ns.  a$  —  yt. 

15.  :«4  +  x1  +  1  by  a;-4  -  x~2  +  1.  x4+l+  a~4. 

16.  a -3  +  a"*  +  1  by  a-*  —  1.  a"1  —  1. 
Divide 

17.  21a;  +  a$  +  a*  +  1  by  3a;*  +  1.        7a;*  -  2a;*  +  1. 

18.  15a— 3a*— 2a-*+8a_1  by  5a*+4.     3a*— Sarb+Var1. 
Find  the  square  root  of 

19.  9a;  -  12a;*  -f  10  —  4a;-*  +  ar*.       3aJ*  -  2  +  .<■-'. 

20.  4aJ"  -f  9a;-"  +  28  —  2Ax~*  —  Km:".       2.r"  -  4  +  ^_^- 


SURDS  —  DEFINITIONS.  261 

SURDS    (RADICALS). 

122.  Surds.  Definitions.  —  When  the  indicated  root  of 
a  quantity  cannot  be  exactly  obtained,  it  is  called  an  irra- 
tional quantity  or  a  Surd. 

Thus,  ^2,  \^i,  Va3,  sja2  +  b2,  a$,  are  surds. 

When  the  indicated  root  can  be  exactly  obtained,  it  is 
called  a  rational  quantity.  Thus  Vx*,  ^9,  Va4,  are  rational 
quantities,  though  in  the  form  of  surds. 

The  order  *  of  a  surd  is  indicated  by  the  index  of  the  root. 
Thus  3y/x,  y/a  are  respectively  surds  of  the  third  and  fifth 
orders. 

The  surds  of  the  most  common  occurrence  are  those  of 
the  second  order ;  they  are  sometimes  called  quadratic  surds. 
Thus  \fs,  Va,  \lx  +  y  are  quadratic  surds.  Surds  of  the 
third  and  fourth  orders  are  called  cubic  and  biquadratic  surds 
respectively. 

When  the  same  root  is  required  to  be  taken,  the  surds  are 
said  to  be  of  the  same  order.  Thus,  Va,  Va  +  &,  and  5?  are 
all  surds  of  the  third  order  or  cubic  surds. 

Surds  are  said  to  be  like  or  similar  when  they  are  of  the 
same  order,  or  can  be  reduced  to  the  same  order,  with  the  same 
quantity  under  the  radical  sign. 

Thus,  4^7  and  5^7  are  like  or  similar  surds  ;  also  53/2  and 
3Vl6  are  like  surds ;  2^3  and  3\/2  are  unlike  surds. 

A  mixed  surd  is  the  product  of  a  rational  factor  and  a 
surd  factor.     Thus  4\/5,  and  3^7  are  mixed  surds. 

When  there  is  no  rational  factor  outside  of  the  radical 
sign,  the  surd  is  said  to  be  entire.  Thus  \f%  and  V^  are 
entire  surds. 

The  rules  for  operating  with  surds  follow  from  the  propo- 
sitions of  the  preceding  Articles  of  this  Chapter. 

*  Sometimes  culled  degree. 


262        TO  REDUCE  AN  ENTIRE   TO  A  MIXED   SURD. 

123.  To  Reduce  a  Rational  Quantity  to  a  Surd 
Form.  —  It  is  often  desirable  to  write  a  rational  quantity  in 
the  form  of  a  surd.     Thus 

a  =y^  =  V«?  =  Vc?!    3  =  yft  =  y/27. 
In  the  same  way  the  form  of  any  surd  may  be  altered. 
Hence  the 

Rule. 

Any  rational  quantity  may  be  expressed  in  the  form  of  a 
surd  of  any  required  order  by  raising  it  to  the  power  corre- 
sponding to  the  root  indicated  by  the  surd,  and  prejixing  the 
radical  sign. 

Examples.  5  =  y/25  =  yl25  =  y5n. 

(a  +  b)  =  \l(a  +  by3  =  V(«  +  b)3. 

124.  To  Introduce  the  Coefficient  of  a  Surd  under 
the  Radical  Sign.  —  We  have 

3^2  =  fi  X  fe  (Art.  123) 

=  y/'J  x  2  [(3)  of  Art,  121]  =  \Zl8. 

2V5  =  yip  X  VIF-  V28  X  5      =  VIo. 

ccy^a  —  x  =  sj'laxr  —  a-3. 
Rule.  Reduce  the  coefficient  to  the  form  of  the  surd  and 
then  multiply  the  surds  together. 


EXAMPLES. 

Express  as  entire  surds. 

1.    11^2.        Aus.  y/242. 

3.   14\':>. 

Ans.  y/'J.sO. 

2.  5  ye.         yiEo. 

4.  cyi 

ysci. 

125.  To  Reduce   an  Entire  to   a  Mixed  Surd. 

We  have 

y/32  =  y/i6  x  2  =  \/T<;  x\/2=  iy/2 ; 
also  yaJb  =  V^"0  X  yi  =  ir\b. 


REDUCTION   OF  SURDS   TO  EQUIVALENT  SURDS.     263 

Rule.  Resolve  the  quantity  under  the  radical  sign  into  two 
factors,  one  of  which  is  the  greatest  perfect  power  correspond- 
ing to  the  root  indicated;  extract  the  required  root  of  this 
factor,  and  prefix  the  result  as  a  coefficient  to  the  indicated 
root  of  the  other. 

When  a  surd  is  so  reduced  that  the  smallest  piossible  integer 
is  left  under  the  radical  sign,  it  is  said  to  be  in  its  simplest 
form.     Thus, 

The  simplest  form  of  \Jl28  =  ^64  x  2  =  8^2, 

EXAMPLES. 

Express  in  the  simplest  form  : 

1.  \/288.        Ans.  12^2.  3.    y/36o~3.  Ans.  6a^a. 

2.  y/1029.  7y/3.  4.    \/27a3b5.  3ab°-\j3ab. 

126.  Reduction  of  Surds  to  Equivalent  Surds.— 

To  reduce  surds  of  different  orders  to  equivalent  surds  of  the 
same  order. 

For  example,  take  y/5  and  y/ll. 

y/5  =    5^  =    5*  =    y~5~3  =  Vl25. 

yn  =  n$  =  lit  =  yn5  =  ym. 

In  general,  let  n\Jam  and  qyjbp  be  two  surds  of  different 
orders.  Then  we  have  to  change  these  into  equivalent  surds 
whose  fractional  exponents  have  the  same  denominator. 

We  can  reduce  both  surds  to  the  order  nq  as  follows : 

yam  =  a?  =  a"*  —  '"y/cr", 

p  p?  

and  \bp  =  bq  =  bqn  =  %/&*». 

Thus  the  equivalent  surds  of  the  same  order  are  "\/«"!9  and 

Rule.  Represent  the  surds  with  fractional  exponents;  re- 
duce these  fractions  to  their  least  common  denominator ;  then 
express  the  resulting  fractional  exponents  with  radical  signs, 


2G4 


ADDITION  AND   SUBTRACTION   OF  SURDS. 


and  reduce  the  expressions  under  the  radical  signs  to  their 
simplest  forms. 

Note.  —  In  this  way,  surds  of  different  orders  may  be  compared. 
Thus,  if  we  wish  to  know  which  is  the  greater,  V^  or  |*Kll,  we  have 
only  to  reduce  them  to  the  same  order,  as  above;  we  see  that  the 
former  is  greater  because  125  is  greater  than  121. 

EXAMPLES. 

Express  as  surds  of  the  twelfth  order,  with  positive  ex- 
ponents : 


Ans.  ly/a;4. 
1 

vS[af 

Express  as  surds  of  the  same  lowest  order : 


1.  xK 

2.  a-1  -:-  ar 

1 

3.  — • 

a  * 


4.    xk  Ans.   \a?. 

5.  yx^r.         is/x^i. 

G.    v&x  V«"ljr"2-     Va 


7.  y/a,V^  Ans.  \/a9,  'y/a™. 

8.  v^v^-        W,W- 


d.^x^y/x10.  Ans.  v^,1^0. 
10.  y/cW,yab.    ^M^y/aW. 


11.  Which  is  the  greater  \/l4  or  \/G?  *6. 

127.  Addition  and  Subtraction  of  Surds.  —  Let  it 

be  required  to  find  the  sum  of  ^12,  ^7o,  -\J±8,  and  fit). 
Here  we  have  (Art.  125) 

\!\2  +  \jTb  -  y/te  +  \J5Q  =  2\/S  +  5\/3  -  4\/3  +  5^2 
=  (2  +  5  -  4)v/3  +  5^2 
=  3\/3  +  5y/2. 
Rule.  .Reduce  the  surds  to  their  simplest  form;  then  add 
or  subtract  the  coefficients  of  similar  surds  and  prefix  the  result 
to  the  common  surd,  and  indicate  the  addition  or  subtraction 

of  unlike  surds.  

Thus,  3y/20  +  4y/5  +  fa  +  y/75 

=  Gy/5  +  40i  +  l\!-i  +  5^3 
-  10^'  +  5^3. 


MULTIPLICATION   OF  SURDS.  2G5 

EXAMPLES. 

Find  the  value  of  the  following : 

1.  3^45  +  7^5  -  y/20.  Ans.  14^5. 

2.  4033  +  5\Il  -  8^28.  V^- 

3.  y/44  -  5\/l76  +  2^99.  -12y/ll. 

4.  2^303  -  5^243  +  y/l92.  -15^3. 

5.  2VI  +  8VS-  3V2- 

g.  y4o  -  iV320  +  V135-  3V5- 

128.  Multiplication  of  Surds. —  (1)  When  the  surds 
are  of  the  same  order. 

To  multiply  a  ny/x  by  b  n\Jy. 

Here     a"y/x  x*  bny/y  =  axn  x  &#"  [Art.  118,  (5)] 
ii  i 

=  abxnyn  =  ab(xy)n  (Art.  121) 

=  ab'yjxy. 

(2)  T7/ie»  the  surds  are  of  different  orders. 

To  multiply  a\x  by  by/y. 

_  _  i  2. 

Here  ayas  X  by/y  =  axn  x  6?/m 

=  abx""'y"">  (Art.  126) 

=  a6(«"y()'""  (Art.  121) 
=  abmylx^Y. 

Rule.  When  the  surds  are  of  the  same  order,  multiply  sep- 
arately the  rational  factors  and  the  irrational  factors.  When 
the  surds  are  of  different  orders,  reduce  them  to  equivalent 
surds  of  the  same  order,  and  proceed  as  before. 

Thus,  3\/2  x  7v/6  =  21^12  =  42^3. 

5^2  x  2^5  =  5V2*X  2^ 

=  loysoo. 

A  compound  surd  is  an  expression  involving  two  or  more 


266  MULTIPLICATION   OF  SURDS  —  EXAMPLES. 

simple  surds.  Thus  203  —  3\Jb,  and  30t  +  V&  are  compound 
surds. 

The  multiplication  of  compound  surds  is  performed  like 
the  multiplication  of  compound  Algebraic  expressions. 

Multiply  2Va;  —  5  by  30c. 

The  product  =  3\/^(20c  -  o)  =  Gx  -  150k. 

Note.  — To  multiply  a  surd  of  the  second  order  by  itself  is  simply 
to  remove  the  radical  sign ;  therefore  \x  x  \Jx  =  x. 

Multiply  603  -  5  0>  by  205  -|-  302. 

The  product  =  (603  -  5y/2)  (203  +  302) 

=  36  +  18  ft  -  1003  -  30  =  6  +  Sy/S. 
The  following  case   of    the   multiplication   of   compound 
surds  deserves  careful  attention.     The  product  of  the  sum 
and  difference  of  any  two  quadratic  surds  is  a  rational  quan- 
tity.    Thus 

(303  +  403)  (303  -  403)  =  (3  ft?  -  (4y/3)2 
=  45  -  48  =  -3. 
Also  (\Ja  +  00(0*"  -  00  =  (ft)2  ~  (ft)2  =  a  -  b. 
A  binomial  in  which  one  or  both  of  the  terms  are  irra- 
tional, is  called  a  binomial  surd. 

When  two  binomial  quadratic  surds  differ  only  in  the  sign 
which  connects  their  terms,  they  are  said  to  be  conjugate. 
Thus 

30*  +  4y/3  is  conjugate  to  3\/5  —  403. 

Similarly,  a  —  0t2  —  b'2  is  conjugate  to  a  +  0r  —  b2. 
The  product  of  two  conjugate  surds  is  always  rational. 

EXAMPLES. 

Find  the  value  of 


1.  2014  x  0>T.     Ans.  1405. 

2.  80$  X  03.  1203. 

3.  2)JTr>  x  80S  3003. 


4.  yi68  x  0147.  Ans.  14*0). 

5.  0*  X  \  2.  V10*- 

6.  V^2xV8xVixVI«        V2- 


TO  RATIONALIZE  THE  DENOMINATOR  OF  A  FRACTION.   267 

7.  (30c  —  5)  X  2y/a\  Ans.  6x  —  100c. 

8.  (V®  —  sja)  x  20c.  2a;  —  20X8. 

9.  (07  +  503)  (2^7  -  403).  60*1  -  46. 

10.  (30i  -  400(203  +  302).  6  +  010. 

11.  (5  +  802)  (5  -  302).  7. 

12.  (30t  +  0b  —  9a)(30»  —  0c  —  9a).  18a  —  x. 

129.  To  Rationalize  the  Denominator  of  a  Frac- 
tion. —  The  process  by  which  surds  are  removed  from  the 
denominator  of  any  fraction  is  known  as  rationalizing  the 
denominator. 

(1)  When  the  denominator  is  a  monomial. 

_2_  _        203      _  203 
03      03  x  03        3 

'2  X  9       l3/l8       0L8 


*W 


3x9        V  27         3 

Rule.    Multiply  both  terms  of  the  fraction  by  any  factor 
which  will  render  the  denominator  rational. 

(2)  When  the  denominator  is  a  binomial  quadratic  surd. 

b2 
Rationalize  the  denominator  of     >  „  ■ 

y/a2  +  b2  +  a 

„,                  .                       b2                  y/a2  +  b2  -  a 
The  expression  =  -== X     , 

y/a2   +   &2  +   a  y/a2   +   &2   _    a 

=  yfy/a"  +  fr2  -  a]  =  ^r^i  _  a< 
(a2  +  &'2)  -  a2 
Rule.    Multiply  both   numerator  and   denominator  of  the 
fraction  by  the  surd  which  is  conjugate  to  the  denominator. 

When  the  denominator  of  the  fraction  is  rationalized,  its 

numerical  value  can  be  more  easily  found,    Thus,  the  numeri- 

-  2 

cal  value  of  f  03  can  be  found  more  easily  than  that  of  "7=* 


268 


DIVISION   OF  SURDS. 


Given 


^5  =  2.2360G8,  find  the  value  of 


29 


2^/5 


It  might  seem  at  first  sight  that  we  must  subtract  twice 
the  square  root  of  5  from  7,  and  divide  29  by  the  remainder 
—  a  troublesome  process,  as  the  divisor  would  have  7  figures. 
We  may  avoid  much  of  this  labor  by  rationalizing  the  de- 
nominator.    Thus, 

29        =  29(7  +  2y/5) 
7  -  2v/5  49  -  20 

=  11.472136. 


=  7  +  2^5 


EXAMPLES. 

Rationalize  the  denominators  of 


V3 

1. 

V» 

2. 

\/f- 

3. 

\Ti- 

4. 

VI- 

5 

2  + 

y/5 

V/5- 

-    1 

Ans.  W15- 

We- 

7  +  3y/5 


7^3  —  5^/2 


iOy/6  -  2y/7 
3\/6  +  2y  7  ' 


8  -  y/42. 

y/7- y/2 


3    y/7  +  y/_2__ 

9  +  2^14  5 

Since  anyjx  x  &>A/  =  ab^Jxy 


130.  Division  of  Surds. 

(Art.  128),  therefore 

abnyjxy  ■+■  an\]x  =  b\y. 
Rule.    When   the   surds   are   of   the   same   order,   divide 
separately   the  rational  factors   and   the   irrational   factors. 
When   the   surds   are   of   different   orders,   reduce    them    to 
equivalent  surds  of  the  same  order,  and  proceed  as  before. 
Then  the  denominator  may  be  rationalized  (Art.  129). 
4^75    =     4  X  5y/3 
25^56        25  X  2^14 


Thus,  4^75  -=-  25y/56  = 

=  «Vft  =  ^l 


X    14 
X   14 


y/42 
35' 


BINOMIAL   SURDS  —  IMPORTANT  PROPOSITIONS.     269 


The  only  case  of  division  of  a  compound  surd  which  we 
shall  consider  is  that  in  which  the  divisor  is  a  binomial 
quadratic  surd.  We  may  express  the  division  by  means  of 
a  fraction,  and  then  practically  effect  the  division  by  ration- 
alizing the  denominator.  Thus, 
Divide  y/3  +  y/2  by  203  -  02. 

y/3_+  v/2_  =    (y/3  +  y/2)  (203  +  y/2) 
203  -  y/2       (203  -  y/2)  (2^3  +  y/2) 

8  +  303  ___  8  +  303 
12  —  2  10 


The  quotient 


EXAMPLES. 
Find  the  value  of 

1.    210384  -*-  80)8. 

J.ns.  3 y/3. 

3.    -I3y/l 

25  -*-  5y/65. 
-4ns.  —  y/13 

2.    5y/27-3y/24.     ^. 

4.    6014  -=- 

-  2021.         y/(5. 

5.    29  -j-  (11  +  307). 

11  -  307 
2 

6.    (3^2  -  1)  -  (302  +  1 

)• 

19  -  602 
17 

7.  (2y/3  +  700  -=-  (5y/3  - 

8.  (2*  -  0^)  -  (20^  - 

-  4y/2). 
y)- 

2  +  03 

131.  Binomial  Surds.    Important  Propositions. 

(1)    The  square  root  of  a  rational  quantity  cannot  be  partly 
rational  and  partly  a  quadratic  surd. 

If  possible,  let  y/a  =  b  +  y/c. 
Squaring,  we  have    a  =  &2  -f-  260;  +  c. 

V  26 

that  is,   a  surd   is   equal    to  a  rational  quantity,   which   is 
impossible. 


270  SQUARE   ROOT   OF  A   BINOMIAL    SURD. 

(2)  In  any  equation  consisting  of  rational  quantities  and 
quadratic  surds,  the  rational  parts  on  cadi  side  are  equal, 
and  als<>  the   irrational  parts. 

If  x  +  V y  =  «  +  \b,  then  will  x  =  a,  and  y  =  b. 

For  if  x  is  not  equal  to  a,  suppose  x  =  a  +  m ; 
then  a  +  m  +  y/?/  =  a  +  y'ft  ; 

that  is,  m  +  \Jy  =  \Jb, 

which  is  impossible  by  (1).     Hence  x  =  a,  and  therefore 
s/y  =  sjb. 

Note.  — When  x  +  \fy  =  a  +  sjb,  we  can  conclude  that  x  =  a  and 
sjy  =  \l>  only  when  \Jy  and  Sjb  are  really  irrational.  We  cannot,  for 
example,  from  the  relation  6  +  V'-A  =  5  +  V^,  conclude  that  6  =  5  and 

s/l  =  sl9. 


(3)  If  >Ja  +  s/b  =  sjx  +  sjy,  then  \a  -  sfb  =  Va:  -  \ly. 
For  by  squaring  the  first  equation  we  have 

a  +  sjb  =  a;  +  y  +  2^. 
a  =  x  +  ?/,  and  yo  =  2\xy. 
Subtracting,  a  —  sjb  =  x  —  2sjxy  -+-  # ; 

.-.     V a  —  \lb  =  \Jx  -  \ly. 

132.  Square  Root  of  a  Binomial  Surd. —  The  square 
root  of  a  binomial  surd,  one  of  whose  terms  is  rational,  may 
sometimes  be  expressed  by  a  binomial,  one  or  each  of  whose 
terms  is  a  quadratic  surd. 

Let  a  +  y  b  be  the  given  binomial  surd. 


Assume  Va  +  \b  =  yx  +  sjy. 

By  (3)  of  Art.  131,     Va  -  \lb  =  sfx  -  sjy.  . 

Multiplying  (1)  by  (2),  ya2  -  b  =  x  -  y.     .  . 

Squaring  (1)  a  +  sjb  =  x  +  2\'.ry  +  y. 

Therefore  by  (2)  of  Art,  181,  a  =  x  +  y.     .  .     , 


(1) 
(2) 
(3) 
(4) 
(5) 


SQUARE  HOOT   OF  A   BINOMIAL   SUED.  271 

Hence,  from  (3)  and   (5),  b}'  addition  and  subtraction,  we 
have 


_  a  +  y  a2 
2~~ 


(6) 


a  —  va2  —  b  ,„. 

y=  — 2 — ' (7) 

which  substituted  for  x  and  y  in  (1)  and  (2)  will  give  the 
values  of  \a  +  s/b  and  \  a  —  ^b. 

Find  the  square  root  of  16  +  2^55. 

Here  a  =  16,  and  ^b  =  2^55. 
Then  a2  —  b  =  256  -  220  =  36, 

which  in  (6)  and  (7)  gives 

x  =  ±(16  +  6)  =  11. 
y  =  |(16  -  6)  =     5. 
Hence  Vl6  +  2055  =  *Jli  +  05. 

From  the  values  of  x  and  y  in  (6)  and  (7),  it  is  clear  that 
each  of  them  is  itself  a  complex  surd  unless  \cr  —  b  is 
rational ;  and  the  expression  \jx  -f  \Jy  will  be  more  compli- 
cated than  \a  +  \lb  itself.  Hence  the  above  method  for 
finding  the  square  root  of  a  +  ^b  fails  entirely  unless  a2  —  b 
is  a  square  number  ;  and  as  this  condition  is  not  often  satis- 
fied, the  process  has  no  great  practical  utility. 

The  square  root  of  a  binomial  surd  may  often  be  found  by 
inspection.  For  we  see  from  (4)  and  (5)  that  we  have  to 
find  two  numbers  whose  sum  is  a  and  whose  product  is  6  ; 
and  if  two  rational  numbers  satisfy  these  conditions,  they 
can  generally  be  found  at  once  by  inspection.     Thus 

1.    Find  the  square  root  of  11  +  2030. 

We  have  only  to  find  two  numbers  whose  sum  is  11,  and 
whose  product  is  30_;  and  these  are,  evidently  6  and  5. 

Hence  11  -f  2030  =  6  +  2^6  X  5  +  5 
_  =  (^6  +  fo)\ 
.'.     y/6  +  V5  =  the  square  root  of  11  +  2030. 


272  EQUATIONS  INVOLVING   SURDS. 

2.    Find  the  square  root  of  53  -  12^10. 

We  must  write  the  binomial  so  that  the  coefficient  of  the 
surd  is  2.     Thus 

53  -  12v/l0  =  53  -  2\/360. 

The  two  numbers  whose  sum  is  53  and  whose  product  is 
3G0  are  45  and  8.  

Hence       53  -  20560  =  45  -  2^45  X  8  +  8 

__  =  (\/45  -  >Jzy. 

.-.     ^53  -  12V/10  =  ^45  -  \J%  =  3\/5  -  2V^2. 

EXAMPLES. 

Find  the  square  root  of 

1.  7  +  2v/l0.  Am.  s/J  +  >J2. 

2.  13  +  2y/30.  )/l0  +  y/3. 

3.  5  +  2^  \Jj  +  l/2. 

4.  47  -  4^33.  2^11  -  0*- 

5.  15  +  2^56.  Y^  +  \ft> 
The  c*i&e  root  of  a  binomial  surd  may  sometimes  be  found 

by  a  method  similar  to  the  one  just  given  for  obtaining  the 
square  root.  But  the  method  is  very  imperfect,  and  is  of  no 
practical  importance. 

133.  Equations  Involving  Surds.  —  Equations  some- 
times occur  in  which  the  unknown  quantity  appears  under 
the  radical  sign.  In  the  solution  of  such  equatious,  special 
artifices  arc  often  required.  We  shall  here  consider  only  a 
few  of  the  easier  cases,  which  reduce  to  simple  equations. 
These  can  generally  be  solved  by  the  following 

Rule.  Transpose  to  one  member  of  the  equation  a  single 
radical  term  so  it  will  stand  by  itself;  then  on  raising  each 
member  to  a  power  of  the  same  degree  as  the  radical,  it  will 
disappear.  If  there  are  still  radical  terms  remaining,  repeat 
the  process  till  all  are  removed. 


EXAMPLES. 


273 


EXAMPLES. 

1.  Solve  2\/x  —  ^Ax  -  11  =  1. 
Transposing,  2\Jx  —  1  =  ^Ax  —  11. 
Squaring,                     Ax  —  A\Jx  +  1  =  Ax  —  11. 
Transposing  and  dividing  by  —  4,  ^x  =  3.         .*.     a;  = 

2.  Solve  V^a?  —  ^1  —  x  +  \/x  = 
Transposing,  \x  —  ^1  —  a  = 
Squaring, 


\Jx. 

2)Jx  +  ». 

4y/cc  +  4a;. 


a;  —  y  1  —  x  = 
Canceling  x  and  squaring,  1  —  x  = 
Transposing  and  squaring,  25a-2  =  16a;. 

Dividing  by  25a;,  x  =  ^§. 

When  radicals  appear  in  a  fractional  form,  the  equation 
should  be  first  cleared  of  fractions  in  the  usual  way  before 
performing  the  involution. 

3.    Solve  6Vfc-ll  =  2V^  +  1. 
3\/x  \jx  +  6 

Clearing  of  fractions 

6a;  +  25y/a;  -  66  =  6x  +  S)Jx. 

.-.     22y/a;  =  66.         .-.     x  =  9. 
Solve  the  following  equations. 

4.  \jx  -  5  ==  3.  Ans.  14. 

5.  SjAx  -  7  =  5.  33. 

6.  y/5a;  -  1  =  2\jx  +  3.  13. 


7.    13  -  y/5a; 


4  =  7, 


203  -  7a;  -  3^8x  -  12 
VI  +  ^3  +  ^6x  =  2. 


44. 
f 

6. 


274 


EXAMPLES. 


EXAMPLES. 


1.    Multiply  a~%  by  a<L 

-4ns.  a". 

2.    Multiply  Zcfib*  by  4a^&i 

12aM 

3.    Divide  x%  by  as*. 

x&. 

4 .    Divide  a  ~  2x  ~  *  by  a  ~~ 3. 

ax~K 

Find  the  numerical  values  of 

5.    1G~K             Ans.  |. 

8. 

4-1 

Ans.  -|. 

6.    27*.                        81. 

9. 

16" 

"-                        i- 

7.    8i.                            4. 

10. 

(t^o 

)-*.                25. 

Express  with  positive  exponents  : 

11.    sl^y  +  y/o?. 

^L?is.  a%"k  +  a%. 

12.    y/'a'2.«5  +  \/ah/. 

a$a$  +  a%*. 

ycc-1         y  a         4  yce    d 

a    ,    1     .    a-'"' 

354  +  —  +  T- 
a*        4 

14.    (a  -j-V^)"  +  V;  x  V *  ~V 

,2 

15.  /^Vs  -  (4SY. 

1 

17. 


is. 


2»  +  : 


1" 


(2")"-1        (2»-1)"  +  1 
Express  with  radical  signs  and  positive  exponents : 


u° 

cK 


19. 


2a~ 


+ 


3a 


_1                   .          2,1,4 
— .  -4na.  —  H ^  +  — • 


20.   a*  X  a*  +  a"* 


Va18  +  Vo*- 


EXAMPLES.  275 

21.    ah-1  -  a~h.  Ans. 


Va?        b 


22.  a-36~i  +  3a5&-l.  _JL  +  31"/^!. 

Multiply 

23.  1  +  x*  +  x$  by  1  -  xi.  1  -  as*. 

24.  a+  —  x%  +  an  —  a;-*  by  a;2  +  aji.  a;2  —  1. 

25.  cc"  +  aj2  4-  1  by  a;-n  +  a;-2  +  1. 

^Lrcs.  xn  +  2aS*  +  3  +  2a;-"  4-  »-« 
2G.    1  -  23}/x  -  2x*  by  1  -  \[x.       1  -  a*  -  2x$  +  2x§. 

27.  2  y/a5  -  a*  -  -  by  2a  -  3^A  -  a~i 

-dws.  4a*  —  8a^  —  5  4-  10«~*  4-  3a~8. 
Divide 

28.  16a"3  4-  6a-2  +  5a"1  -  G  by  2a"1  -  1. 

Ans.    8a-2  4-  7a-1  4-  6. 

29.  21ato  +  20  -  27a*  -  26a2*  by  8a*  -  5. 

Ans.  lcrx  4-  3a*  —  4. 

30.  8c- ■  -  8c"  +  5c3"  -  3c -3"  by  5c"  -  3c-". 

Ans.  c-n  —  1  4-  c~"n. 

31.  1  -  y/a  -  2a  4-  2a2  by  1  -  a*.  1  -  2a  -  2ai 

32.  ^  -  ^  4-  afy  —  yi  by  a;*  —  yK  x  +  y. 

33.  a^  +  6^  -  c*  +  2a*6*  by  a*  4-  6^  +  ci       a*  +  &£  -  ci 
Find  the  values  of 

35.    (a^^  +  a-^)3.       a»-34-3a'.«-14-3a-^  +  a-1a;3. 
3G.    (|a±-  a"^)2.  i«S  _  f  +  a-l 

Q7  a^  —  8cdb  1,1        „  .  x 


«»  +  2y/a&  +  4P 

a;  —  7a;? 
x  —  Sy/a;  —  14 


+(i+sr 


276 


EXAMPLES. 


Find  the  square  root  of 

39.    4aAi~2  -  12XU-1  +  25 


K). 


4a;«  +  Ax  -f  2x* 


-  24x~\t  +  16a;-2 a2. 

Ans.  2a'a_1  —  3  +  4a?-1a. 
4a:*  +  a?i       05*  —  2a?*  4-  a?i 


41.    4a  -  12a*6*  +  9&t  +  16a*c 

_  246 M  +  16c*. 

Express  as  entire  surds 

^?<s.  2a*  -  36*  +  4ci 

42.    ?«y  2^2.    ^  ^ 
2c  V  9a2&                v 

44 

at/ .    Ans.  \la2b'. 

V  a""2                v 

.0     2a.  3/2  7a?4              3/3 — 

43.    — 1/ \8ax. 

3a?V    a2                v 

45 

a?"V     y*                   Xy 

Express  in  the  simplest  form  : 

46.    3^150  and  2^720. 

Ans.  1503  and  2403. 

47.    V-108o?y. 

—  3xy  y/4a?. 

48.    sja*  +  2a2b  +  ab\ 

(a  +  &)\/a. 

49.    y/Sx^j  -  24x*y*  +  24a?y  - 

8.t^4.          2(a  -  y)V^- 

50.    ^768,  Vl250,  y/3888. 

403,  5*^2,  6*03. 

Express  as  surds  of  the  same  lowest  order : 

51.    y/o^2,  "VaV. 

^is.  "V'Ws  "y^r4. 

52.    y/^Vll,  Vl3. 

Vl25,  Vl21,  Vl8. 

53.   ^8,  ^3,  Vc. 

V64,  y*i,  y&. 

54.    3*,  2s  5*.                                 3 

y/6561,  ^512,  ^15625. 

Find  the  value  of 

55.    v/^43  +  y/27  +  \/48. 

Ans.  1603. 

56.    2  y/189  +  3y«75  -  7^/56. 

r-3/^ 

7^7. 

57.    5*^81  -  7*^192  +  4y/648. 

11  Vs. 

58.    3  y/162  -  7*0*2  +  V1250- 

0. 

59.    3\/2  +  4^8  -  032. 

702. 

60.    2y/4  +  5*032  -  0108. 

9\/4- 

61.   4^128  +  4075  -  5y/T 

12. 

2003  -  130>. 

EXAMPLES. 

277 

62. 

;~>yi28  x  2y/432. 

^ws.  240  \/4. 

63. 

3  y  6  x  4  y/«. 

12  'y/aW. 

64. 

y/2  x  y§  x  v*- 

Jy/648000. 

65. 

4y/5  x  2'Vii. 

8V15125. 

66. 

2^*V£ 

2y/2 
a 

67. 

(2y/3  +  3y/2)2. 

a;  — 

30  +  1203. 

68. 

(\/a?  +  V-c-  1)V*-  !• 

1  +  y/a;2  -  a;. 

69. 

(y/a;  +  a  —  y^a;  —  a)  y/a;  +  a. 

x  +  < 

%  —  y/a;2  —  a2. 

70. 

(y/2  +  y/3  -  y/5)  (y/2  +  0*  + 

y/5). 

2y/6. 

71. 

(Vl2  +  v/l9)(Vl2  -  Vl9)- 

5. 

72. 

(a,2  +  a;V/2  +  l)(a»-  ^2  +  : 

0- 

a;4  4-  1. 

Rationalize  the  denominator  of 

73. 

203  +  3y/2 
5  +  2y/6  ' 

^4.Wi 

?.  30>  -  203. 

74. 

2  y/5 
V/5  +  03* 

5  -  y/T5. 

7."). 

3+03 

3  -  y/5' 

i(7  +  303). 

76.    ■'  x  —  0a:2  -  if. 

x  +  y/^  _  ^2 


77.  ,■  •  va;2  +  a2  —  a. 

y/a?  +  a2  +  a  v      T 

78.  y/l  +  x2  -  y/1  -  a^  i  _  y/F~4 
y/l  4-  a:2  4-  y/l  -  a;2'  ^ 

Find  the  numerical  value  of  the  following  to  five  places  of 
decimals  : 

79.  It,   i£.  Ans.  9.89957,  3.77964. 

y/2    y/7 

80.  t?,   — L.  ,  .81649,  .28867. 
^3     203 


278  EXAMPLES. 

gl     y/5  +  Vj^  Ans   ^  _  ^  =  .50402. 

4  +  y/l5 

82.     V*  ~  2_.  y/5  +  2  =  4.23607. 

9  -  4y/5 

Find  the  value  of 

83.  sy/n  +    *  {ft. 

2y/98       7\/22 
84.    l^^Gj/JM  ^ 

5y/ll2        y/392 


85.    (3  +  y/5)  (y/5  -  2)  -=-  (5  -  ^5) 
86 


5 


y/«      _^_  \/a  +  y/g  y/«a 

y/a  —  y/sc  y/x  a  —  x 

87>   2y/l5  +_8  _^  8y/3  -  6y/5.  4  +  ^ 
5  +  y/l5        5y/3  —  3y/5 

1  +  L=.  2a-. 


a;  +  y/a2  —  1        x  —  y/ar 


89     ^  +  *  +  ^  -_1  +  V**  +  1  ~  Vglzii.        2a2 

yV  +  i  _  y/a-a  _  i  y/^  +  1  +  y/a;2  -  1 

90.  _ »  +  ^  y/3. 

5y/3  -  2y/l2  -  y/32  +  y/50 

Find  the  square  root  of 

91.  41  -  24y/2.  -dns.  402  -  8. 

92.  83  +  12035.  2y/5  +  3^7. 

93.  101  -  28013.  2yi3  -  7. 

94.  117  -  36y/io.  6^2  -  30i. 

95.  280  +  56^21.  14  +  2y/21. 

96.  8  +  40}.  05  +  y/2. 

97.  4  -  y/15.  y/f  -  y/f- 

98.    75  -  12021.  3y/7  -  20J. 


EXAMPLES.  279 

99.    16  +  5y/7.  V^¥  +  V^i- 

100.  6  +  2y/2  +  20*  +  205.  1  +  0>  +  0*- 

101.  5  +  y^lO  -  ^6  -  ^15.  1  +  Vf  -  Vt- 
Solve  the  following  equations  : 

102.  S  -  20b  =  4.  Ans.  4. 

103.  G  +  0b  =  2^/12  +  x.  4. 

104.  \Jx  -  3  -  0c  +  12  =  -3.  4. 

105.  y/3.K  +  10  -  \j3x  +  25  +  3  =  0.  -3. 
91 


10G.    y/3a:  -f  ^3x  +  13  =  12 

y/3a;  -f-  13 


107.    2  +ya;  -  5  =  13.  1336 


108.    0J  +  2x  -  \j2x 


y/9  +  2x 


109.    0»  -  1  +  V^  =  -7=-  f 

no.   05  -  0>  -  8  =     _2     .  9 

0c  -  8 

111.  0»  +  11  -  0c  =  1-  25 

112.  03a;  +  4  +  y/3a;  -  5  =  9.  7 


113.  «0  -  a-  =  &0i  -  x.  Ans.  -~^— 

v  v  a  +  b 

114.  0»  +  13  -  0b  -  11  =  2.  36 

115.  ^~  1  =  1  +  V^-  \  3 
0jaj  +  1                      2 

a(6  4-  c) 


16. 


V  a  +  a        V  a  —  a        V  a2  —  x2 


b  -  c 


117.  vVa;  +  3  -  */\/x  -  3  =  V^20b.  9 

118.  y/(l+a)2+(l—  a)a;+y/(l—  a)3+(l+a)«=2a.       8 

no     ./ — I —  ,  ./ — TT      ./"       a2+63+c2—  2a&— 26c— 2ca 

119.  Va+a  +  va;+&  =  V^'-       — ! ! 

4c 


280  PURE    QUADRATIC  EQUATIONS. 


CHAPTER     XIV. 

QUADRATIC  EQUATIONS  OF  ONE  UNKNOWN 
QUANTITY. 

134.  Quadratic  Equations.  —  An  equation  which  con- 
tains the  square  of  the  unknown  quantity,  but  no  higher 
power,  is  called  a  quadratic  equation,  or  an  equation  of  the 
second  degree. 

A  Pure  quadratic  equation  is  one  which  contains  only  the 
square  of  the  unknown  quantity  ;  it  is  sometimes  called  an 
incomplete  quadratic  equation.  An  Adfected,  or  Affected,* 
quadratic  equation  is  one  which  contains  both  the  square  and 
the  first  power  of  the  unknown  quantity  ;  it  is  also  called  a 
complete  quadratic  equation. 

Thus,  2x2  =  50,  and  ax2  +  6  =  0, 

are  pure  quadratic  equations  ;  and 

2x2  —  hx  —  4,  and  ax2  +  bx  +  c  =  0, 
are  affected  quadratic  equations. 

135.  Pure  Quadratic  Equations.  —  A  pure  quadratic 
may  be  solved  for  the  square  of  the  unknown  quantity  by 
the  method  of  solving  a  simple  equation. 

Let  it  be  required  to  solve 

x2  —  13    .    a;2  —  5        n 

4- =  b. 

3  10 

Clear  of  fractions, 

10.tr  -  130  +  3a;2  -  15  =  180. 

...  iSx1  =  325. 

,                          .-.  aja  =  2'). 

Extracting  the  square  root  x  =  ±5. 

•  The  term  adfected,  or  affectt  </,  was  introduced  by  Vleta,  about  the  year  1600,  to 
distinguish  equations  which  Involve,  or  arc  affected  with,  different  powers  of  the 

unknown  quantity  from  those  which  contain  one  power  only. 


PURE   QUADRATIC   EQUATIONS.  281 

Iii  this  example  we  find  that  x2  =  25.  Therefore  x  must 
be  such  a  number  that  if  multiplied  by  itself  the  product  is 
25;  i.e.,  x  must  be  the  square  root  of  25;  we  prefix  the 
double  sign  to  5  because  the  square  root  of  a  quantity  may 
be  either  positive  or  negative.     [Art.  Ill,  (1)]. 

Note. —  In  extracting  the  square  root  of  the  two  members  of 
the  equation  x2  =  25,  it  might  seem  at  first  that  we  ought  to  prefix  the 
double  sign  to  the  square  root  of  each  side,  and  write  ±x  =  ±5.  An 
examination  however  of  the  various  cases  shows  this  to  be  unneces- 
sary, because  we  obtain  no  new  results  in  so  doing.  Thus,  if  we 
write  ±x  =  ±5,  we  have  the  four  cases: 

+x  =  +5,  -fac  =  —5,  —x  =  +5,  —x  =  —5; 

but  the  last  two  are  equivalent  to  the  first  two,  and  become  identical 
with  them  on  changing  the  signs.  Hence  there  are  no  new  results 
obtained,  and  therefore  when  we  extract  the  square  root  of  the  two 
members  of  an  equation,  it  is  sufficient  to  put  the  double  sign  before 
one  member  only.     Thus  the  equation  has  two  roots,  and  only  two. 

A  pure  quadratic  equation  can  always  be  reduced  to  the 

form 

ax2  +  6  =  0; 

for  all  the  terms  containing  x2  may  be  reduced  to  one  term, 
as  ax2 ;  and  the  known  terms  to  another,  as  b. 

By  transposing  b,  dividing  by  a,  and  putting  q  =  — ,  the 

equation  may  be  written 

x2  =  q. 

Such  an  equation  is  called  a  binomial  equation,  because  it 
has  but  two  terms. 

Solving  this  equation  by  extracting  the  square  root  of  each 
member,  we  have 

x  =  ±vV 

That  is,  Every  pure  quadratic  equation  has  two  roots, 
numerically  equal,  but  ivith  contrary  sign*. 

Hence,  for  the  solution  of  a  pure  quadratic  equation  we 
have  the  following 


282  AFFECTED    QUADRATIC   EQUATIONS. 

Rule. 
Find  the  value  of  the  square  of  the  unknown  quantity  by  the 
rule  for  solving  a  simple  equation,  and  then  extract  the  square 
root  of  both  members. 

EXAMPLES. 

Solve 

1.  llx-2  -  44  =  bx2  +  10.  Ana.  x  =  ±   3. 

2.  (x  +  2)'2  =  4x  +  5.  x=  ±   1. 


4.    14  -  vV2  -  36  =  6.  aj=   ±10. 

136.  Affected  Quadratic  Equations.  —  An  affected 
quadratic  equation  can  always  be  reduced  to  the  form 

ax2  +  bx  +  c  =  0 ; 

for  all  the  terms  containing  x2  may  be  reduced  to  one  term, 
us  ax2;  those  containing  x  to  one,  as  bx;  and  the  known 
terms  to  another,  as  c. 

If  we  divide  by  a,  and  \mt  p  =  -,  and  o  =  —  - -,  the  equa- 
a  a 

tion  may  be  written 

x2  +  px  =  q, 

where  jt  and  q  are  positive  or  negative.  This  is  called  the 
General  Quadratic  Equal  inn. 

Let  it  be  required  to  solve  this  equation.  If  the  first 
member  of  this  equation  were  a  perfect  square,  we  might 
solve  it  by  extracting  the  square  root,  as  in  Art.  135.  To 
ascertain  what  must  be  done  to  make  the  first  member  a  per- 
fect square,  let  us  compare  it  with  the  square  of  the  binomial, 

x  +  *-,  which  is  a;'2  -f  px  +  ^-. 

Thus,  we  see  that  x2  +  px  is  rendered  a  perfect  square  by 
the  addition  of  *-  ;  i.e.,  hi/  the  ad, in  inn  of  tine  square  of  half 


AFFECTED    QUADRATIC  EQUATIONS.  283 


the  coefficient  of  x.      Hence,  adding  *-  to  both  members,  to 

4 

preserve  the  equality,  we  have 

X2  +  px  +  £-  =    q  +  ?-. 
4  4 

This  is  called  completing  the  square. 

Extracting  the  square  root  of  each  member,  we  have 


.+f-±v/«+£ 


..-l±v/g+f 


(i) 


— -f+v/9+£--f-y«+£ 

Thus  there  are  too  roots  of  a  quadratic  equation. 

Note  1.  —  When  an  expression  is  a  perfect  square,  the  square  terms 
are  always  positive.  Hence,  the  coefficient  of  xa  must  be  made  +1,  if 
necessary,  before  completing  the  square. 

1.  Solve  a;2  +  6a;  =  27. 

Here  half  the  coefficient  of  x  is  3  ;  add  32, 
x2  +  6a:  +  32  =  27  +  9  =  36. 
Extracting  the  square  root, 

x  +  3  =  ±6. 
.-.     x  =  —  3  ±  6  =  3,  or  -9. 
We  may  verify  these  values  as  follows  : 
Putting  3  for  a;  in  the  given  equation,         9  +  18  =  27. 
Putting  —9  for  x  in  the  given  equation,  81  —  54  =  27. 
These  results  being  identical,  the  values  of  x  are  verified. 
It  will  be  well  for  the  student  thus  to  verify  his  results. 

2.  Solve  7a;  =  x2  —  8. 

Transposing,  so  that  the  terms  which  involve  x  are  alone 
in  the  first  member,  and  the  coefficient  of  x2  is  + 1  ?  we  have 
x2  —  7x  =  8. 


284  AFFECTED    QUADRATIC  EQUATIONS. 

Here  half  the  coefficient  of  x  is  —\\ 
completing  the  square, 

x°-  -  7x  +  (|)a  =  8  +  V-  =  ¥• 
.-.     x  -  f  =   ±f. 

.-.     ®  =  f  ±  f  =  8,  or-1. 
Note  2.  — We  indicate  (J)'2  in  the  first  member. 

3.  Solve  32  -  3x2  =  10a. 

Transposing,  changing  signs,  and  dividing  by  3,  so  as  to 
make  the  coefficient  of  ar  unit}'  and  positive, 

x*  +  lgx=  ^; 
completing  the  square, 

x*  +  ¥<*  +  (f)'2  =  ¥  +  ¥  =  H1- 
.-.    x  +  §  =  ±\i. 

.-.     x  =  -f  ±  V  =  2,  or  -of 
Note  3.  — We  add  (3 )2  and  not  (if-)2,  to  complete  the  square. 

4.  Solve  5a;2  +  llx  =  12. 
Dividing  by  5,  x2  +  *£x  =  ^ ; 

completing  the  square, 

*3  +  ¥*  +  (H)2  =  ¥  +  *3*  =  «*. 

o»_l_    11    +19 

o   .       0/   -f-   T(J   —    X  r<j. 

,>.    11     +     19    —    4      m.    Q 

.    .        X   —     —10     ^    TO    —    IT'    °'  «*• 

Hence,  for  solving  affected  quadratic  equations,  we  have 

the 

Rule  I. 

Reduce  the  equation  so  that  the  terms  involving  the  unknoion 
quantity  are  alone  in  one  member,  and  the  coefficient  of  x2 
is  +1  ;  complete  the  square  by  adding  to  each  member  of 
the  equation  the  square  of  half  the  coefficient  of  x;  extract  the 
square  root  of  both  members,  and  solve  the  resit/Zing  simple 
equation. 

Note  4. — There  are  other  ways  of  completing  the  square  of  an 
affected  quadratic,  which  arc  convenient  in  special  cases,  and  some  of 

which  will  be  given  as  we  proceed;  hut  the  method  just  explained  is 
the  most  important,  and  will  solve  every  case. 


AFFECTED    QUADRATIC  EQUATIONS.  285 

Instead  of  going  through  the  process  of  completing  the 
square  in  every  particular  example,  it  is  more  convenient  to 
apply  the  following  rule  deduced  from  formula  (1)  of  this 
Article : 

Rule  II. 

Reduce  the  equation  to  the  general  form,  x2  +  px  =  q. 
Then  the  value  of  x  is  half  the  coefficient  of  the  first  power 
of  x  ivith  a  contrary  sign,  plus  or  minus  the  square  root  of 
the  second  member  increased  by  the  square  of  half  the  same 
coefficient. 

Note  5. —The  student  should  use  this  method  in  practice,  and 
become  familiar  with  it,  but  at  the  same  time  be  careful  that  he  does 
not  lose  sight  of  the  complete  method. 

5.  Solve  36a;  -  3.x-2  =  105. 

Transposing,  changing  signs,  and  dividing  by  3, 
x2  -  12a;  =  -35. 
Therefore  by  Rule  II,        x  =  6  ±  \Z-35  +  36  =  1. 
.-.     x  =  6  ±  1  =  7,  or  5. 

6.  Solve  Sx~  2  =  — 2. 

2x  -  3       x  +  4 

c,.      ,.,  .        3a;  —  2       3a;  —  8 

Simplifying, =  . 

1     J    °    2x  -  3        x  +  4 

Clearing  of  fractions, 

3x2  -\-  lOx  —  8  =  Gx2  —  25.v  +  24. 

Reducing,      x2  —  ^-x  —  —  -3g2-. 

Therefore,  Rule  II,    x  =  %5-  ±  V7- 

.-.     x  =  5/  ±  2.9-=  lOf,  or 

7.  Solve  x2  —  4x  =  1. 


Rule  II,  x  =  2  ±  ^1+4  =  5. 

.-.     x  =  2  ±  2.236  =  4.236,  or  -0.236. 
These  values  of   x  are   correct   only  to   three    places   of 
decimals,  and  neither  of  them  will  be  found  to  satisfy  the 
equation  exactly. 


286  CONDITION  FOR  EQUAL   ROOTS. 

If  the  numerical  values  of  the  unknown  quantity  are  not 
required,  it  is  usual  to  leave  the  roots  in  the  form 

2  +  Vo,  and  2  -  V5. 
8.    Solve  x2  -  10.x  =   -32. 


Rule  II,  x  =  5  ±  V-32  +  25  =  -7. 

.-.     x  =  5  ±  ^7. 

But  —  7  has  no  square  root,  either  exact  or  approximate 
(Art.  Ill)  ;  so  that  no  real  value  of  x  can  be  found  to 
satisfy  the  given  equation.  In  such  a  case  the  quadratic 
equation  has  no  real  roots  ;  the  roots  are  said  to  be  imagin- 
ary or  imjjossible. 

In  the  examples  hitherto  considered,  the  quadratic  equa- 
tions have  had  two  different  roots.  Sometimes  however, 
there  is  only  one  root.  Take,  for  example,  the  equation, 
x2  —  10a;  +  25  =  0  ;  by  extracting  the  square  root  we  have 
x  —  5  =  0  ;  therefore  x  =  5.  It  is  found  convenient  how- 
ever in  this  and  similar  cases  to  say  that  the  quadratic  has 
two  equal  roots. 

EXAMPLES. 

Solve 

9.  x2  =  x  +  72. 

10.  9x  -  x2  +  220  =  0. 

11.  x2  -  fee  =  32. 

12.  l£x  =  £  -  x2. 

13.  5a;2  =  8*  +  21. 

14.  ^L±J  =  3a;  4-  2. 

a;  —  1 

1K     Sx  -  8       5a;  -  2 

15.    —  =  -. 

x  —  2         x  +  5 

137.  Condition  for  Equal  Roots.  —  To  find  the  rela- 
tion that  must  exist  between  the  knoum  quantities  of  a  quadratic 
equation  in  order  thai  the  two  roots  may  be  equal. 

Take  the  general  equation 

ax2  +  bx  +  c  =  0. 


Ans.  9, 

-  8. 

20, 

-11. 

6, 

-¥• 

b 

-   4. 

3, 

-   b 

3, 

-    1. 

4, 

V- 

CONDITION  FOR   EQUAL   ROOTS.  287 


Transpose  c 

x2  +     .c 
a 

and  divide  by  ( 
c 
a 

-6  ±  v^2  - 

X, 

Rule  II,      x 

X 

a 
-  4ac 

ft2 

4a2 

_  &2 

—  4ac 

4a- 

Denoting  the  root  corresponding  to  the  positive  surd  by 
xv  and  that  corresponding  to  the  negative  surd  by  x2,  we 
have 


-b  +  w  - 

-  4ac 

2a 

-6  -  y/b*  - 

-  4ac 

2a 
Now  we  see  that  if  b2  —  4ac  =  0,  these  two   roots   are 

equal,  and  each  of  them  is . 

2a 

Hence  the  relation,  b2  —  iac  =  0,  is  the  condition  that  the 
two  roots  of  the  equation  ax2  +  bx  +  c  =  0  may  be  equal. 

The  two  roots  are  real  and  unequal  if  b2  —  4ac  is  positive, 
i.e.,  if  b2  is  Algebraically  greater  than  4ac. 

The  two  roots  are  imaginary  if  b2  —  4ac  is  negative,  i.e., 
if  b2  is  Algebraically  less  than  4ac. 

Hence  the  two  roots  of  this  equation  are  real  and  unequal, 
equal,  or  imaginary,  according  as  b'2  is  greater  than,  equal  to, 
or  less  than  4ac. 

Note  1.  —  If  either  of  the  roots  of  a  quadratic  equation  is  imaginary, 
they  are  both  imaginary. 

By  applying  these  tests,  the  nature  of  the  roots  of  any 
quadratic  may  be  determined  without  solving  the  equation. 

1.  Show  that  the  equation  2x2  —  6x  +  7  =  0  cannot  be 
satisfied  by  any  real  values  of  x. 

Here  a  =  2,  b  =  —  6,  c  =  7. 

.-.     b2  -  Aac  =  36  -  4 .  2  •  7  =  -20. 
Hence  the  roots  are  imaginary. 


288     HINDOO   METHOD    OF   COMPLETING   THE  SQUARE. 

Determine  the  nature  of  the  roots  of 

2.  .c-  +  Bx  +1  =  0.  Ans.  Real  and  surds. 

3.  3.r  -  4a;  -4  =  0.  Rational. 

4.  If  the  equation  /-  -\-  2(k  -f-  2)x  +  [)/,•  =  0  has  equal 
roots,  find  k.  Ans.  h  =  4,  or  1. 

When  an  equation  is  in  the  general  form  o.r-  -j-  bx  +  c  =  0, 
instead  of  solving  it  by  either  of  the  rules  in  Art.  lo6,  we 
may  make  use  of  formula  (1)  above  as  follows  : 

5.  Solve  bx2  +  11a-  =  12. 

Here  a  =  5,  b  =  11,  c  =  —12;  substituting  these  values 
in(l) 


x_  -11  ±  V'(ll)2- 

4-5(- 

12) 

10 

-li  ±  v^sTTi 

-11  ± 

19 

_    4 

If' 

or  —3, 

10 

10 

lieh  agrees  with  the  solution 

of  Ex 

4, 

(Ai 

t.  136). 

Solve  by  this  method  the  following 

G.    3a:2  =  15  -  4a;. 

Ans.  f , 

-3. 

7.    2.r-  4-  7x  =15. 

1. 

—  5. 

.s.    5a;2  4-  4  4-  21a;  =  0. 

—i, 

1 

l>  * 

!).    8a;2  =  x  +  7. 

1, 

_7 
8- 

Id.    35  4-  9a;  -  2x*  =  0. 

7, 

2 ' 

Note  2. — Though  we  can  always  find  the  roots  of  a  given  quad- 
ratic equation  by  substituting  in  formula  (1 ),  yet  tbe  student  is  advised 
to  solve  each  separate  equation  either  by  the  method  given  in  Art.  136, 
and  embodied  in  Rule  II,  or  by  one  of  the  two  following. 

138.  Hindoo  Method  of  Completing  the  Square. 
—  When  a  quadratic  equation  appears  in  the  general  form 
ax2  4-  bx  4-  c  =  0,  the  first  member  may  be  made  a  complete 
square,  without  dividing  by  the  coefficient  of  •'"'.  thus  avoid- 
ing fractions,  by  another  method  (called  the  Hindoo  method), 
as  follows : 

Transpose  c,  and  multiply  by  4a, 

4a-.r  4-  lubx  —   —  4ac. 


HINDOO   METHOD    OF   COMPLETING    THE   SQUARE.      289 

Now  since  the  middle  term  of  :my  trinomial  square  is 
twice  the  product  of  the  square  roots  of  the  other  two  (Art. 
41),  the  square  root  of  the  third  term  must  be  equal  to  the 
second  term  divided  by  twice  the  square  root  of  the  first 
term.  Hence,  dividing  Aabx  by  twice  the  square  root  of 
4a2ic2,  i.e.,  by  4acc,  and  adding  the  square  of  the  quotient, 
&2,  to  both  members,  the  first  becomes  a  perfect  square. 

Thus, 

4a'2.e2  +  iabx  +  b2  =  b2  —  4ac. 

Extracting  the  square  root, 


2ax  +  b  =   ±  vV  —  4ac. 


—  b  ±  vV  —  4ac 

.°.     x  = , 

2a 

which  are  the  same  values  we  obtained  in  (1)  of  Art.  137. 

Rule. 

Reduce  the  equation  to  the  form  ax2  -f  bx  +  c  =  0.  Mul- 
tiply it  by  four  times  the  coefficient  of  x2 ;  add  to  each  member 
the  square  of  the  coefficient  of  x  in  the  given  equation;  extract 
the  square  root  of  both  members,  and  solve  the  resulting  simple 
equation. 

Note.  —  This  method  may  be  used  to  advantage  when  Ave  wish  to 
avoid  fractions  in  completing  the  square,  and  it  is  often  preferred  in 
solving  literal  equations.     (See  Note  4  of  Art.  13G.) 

1.    Solve  2x2  —  ox  =  3. 

Multiply  by  four  times  2,  or  8, 

16a2  -  40a  =  24. 

Add  to' each  side  52,  or  25, 

16.«2  -  40.»  +  25  =  49. 

Extract  the  square  root, 

Ax  -  5  =   ±7. 

...     x  =  L±J  =  3,  or  -a 


290  SOLVING   A    QUADRATIC    BY   FACTORING. 

Solve  by  the  Hindoo  method  the  following: 

2.  3a?2  +  5a;  =  2.  Am.  £,  -2. 

3.  63J2  -  12  =  x.  H,  -1|. 

4.  3ar  -f  2a;  =  85.  5,  -of. 

5.  ar.r  —  bcx  +  odte  =  6rf.  -,  — • 

a       c 

139.  Solving  a  Quadratic  by  Factoring.  —  There  is 
still  one  method  of  solving  a  quadratic  which  is  often  shorter 
than  either  of  the  methods  already  given. 

1.  Consider  the  equation  x1  —  2x  —  .15  =  0. 
Resolving  this  into  factors  (Art.  65),  we  have 

(x  -  5)  (x  +  3)  =  0. 

Now  it  is  clear  that  a  product  is  zero  when  any  one  of  its 
factors  is  zero ;  and  it  is  also  clear  that  no  product  can  be 
zero  unless  one  of  the  factors  is  zero.  Thus  ab  is  zero  if  a 
is  zero,  or  if  b  is  zero  ;  and,  if  we  know  that  ab  is  zero,  we 
know  that  either  a  or  b  must  be  zero  ;  and  so  on  for  any 
number  of  factors. 

Similarly  the  product  (a-  -  5)  (x  +  3)  is  zero,  when  either 
of  the  factors,  x  —  5,  x  -f  3,  is  zero,  and  in  no  other  case. 

Hence  the  equation 

(x  -  5)  (x  +  3)\=  0, 
is  satisfied  if  x  -5  =  0,  or  if  a-  +  3  =  0  ;  i.e.,  if  x  =  5, 
or  if  x  =  —3,  and  in  no  other  case. 

Therefore  the  roots  of  the  equation  are  5,  and  —3. 

2.  Solve  x2  —  5x  +  6  =  0. 
Resolving  this  into  factors,  we  have 

(x  -  2)  (x  -  3)  =  0. 

The  first  member  is  zero  either  when  x  —  2  =  0,  or  when 
x  —  3  =  0;  and  in  no  other  case.  Hence  the  equation  is 
satisfied  by  x  =  2,  or  3  ;  and  by  no  other  values  ;  thus,  2 
and  3  are  the  roots  of  the  equation. 

From  these  examples  it  appears  that  when  a  quadratic 
equation  has  been  simplified,  and  lias  all  its  terms   in  the 


SOLVING  A    QUADRATIC  BY  FACTORING.  291 

first  member,  its  solution  can  alwaj's  be  readily  obtained  if 
the  expression  can  be  resolved  into  factors.  Hence  for  the 
solution  of  such  an  equation,  we  have  the  following 

Rule. 

Reduce  the  equation  to  its  simplest  form,  with  all  its  terms 
in  the  first  member ;  then  resolve  the  ivhole  expression  into 
factors,  and  the  values  obtained  by  equating  each  of  these 
factors  separately  to  zero  will  be  the  required  roots. 

3.  Solve  x2  —  Ax  =  0. 
Factoring,  we  have  x(x  —  4)  =  0. 

The  equation  is  satisfied  if  x  =  0,  or  if  x  —  4  =  0,  and 
in  no  other  case.     Hence  we  must  have  either 
x  =  0,  or  x  —  4  =  0. 
.-.     x  =  0,  or  4. 

Note  1.  —  In  this  example  we  might  have  divided  the  given 
equation  by  x  and  obtained  the  simple  equation  x  —  4  =  0,  whence 
x  =  4,  which  is  one  of  the  solutions.  But  the  student  must  be  partic- 
ularly careful  to  notice  that  whenever  we  divide  every  term  of  an 
equation  by  x,  it  must  not  be  neglected,  since  the  equation  is  satisfied 
by  x  =  0,  which  is  therefore  one  of  the  roots. 

Note  2. — When  the  factors  can  be  written  down  by  inspection, 
the  student  should  always  solve  the  example  in  this  way,  as  he  will 
thus  save  himself  a  great  deal  of  unnecessary  work. 

Solve  the  following  by  resolution  into  factors: 

4.  (3a;  -  1)(3»  +  1)  =  0.  Ans.  ±a. 

5.  x2  -  lice  =  0.  0,  11. 

6.  x2  -  3a;  +  2  =  0.  1,  2. 

7.  x2  -  2x  =  8.  4,  -2. 
'    8.    x2  -  2ax  +  4a6  =  26a?.  2a,  26. 

9.    x2  -  2ax  +  8x  =  16a.  2a,  -8. 

Note  3.  —  When  the  student  cannot  factor  the  equation  readily  by 
inspection,  he  should  solve  it  by  Rule  II,  Art.  136,  or  by  Art.  138. 


292    TO  FORM  A   QUADRATIC  WHEN  THE  ROOTS  ARE  GIVEN. 

140.  To  Form  a  Quadratic  when  the  Roots  are 
Given.  —  We  have  seen  (Art.  139)  that  if 

x2  +  px  +  q  =  (as  —  a)  (as  —  b) , 
then  a  and  b  are  the  roots  of  the  equation 

rf+px  +  q  =  0 (1) 

Conversely,  if  a  and  b  are  roots  of  (1),  then  x  —  a  and 
a;  —  b  are  factors  of  the  expression  as2  -f  pa:  +  </,  which 
may  be  proved  as  follows  : 

Since  a  is  a  root  of   (1),  we  have 

a-  +  pa  +  q  =  0 (2) 

Hence  x2  +  pa;  +  q  =  x2  +  px  +  q  —  {a2  +  pa  +  g) 
=  (x  —  a)  (:«  +  a  +  p) . 
.-.     a:  —  a  is  a  factor  of  x2  +  pre  +  </• 
Hence,  if  a  is   a   root   of  (1),  x  —  a  will   be   a   factor  of 
x2  +  pa;  +  7. 

Similarly  it  may  be  shown  that  if  &  is  a  root  of  (1),  then 
x  —  b  will  be  a  factor  of  x2  +  px  -f  7. 

Now  x'2  -\-  px  -\-  q  cannot  have  more  than  two  factors  of 
the  form  x  —  a,  for  the  product  of  any  number  of  factors 
of  the  form  x  —  a  must  be  of  the  same  degree  in  x  as  the 
number  of  the  factors  ;  also  a;2  -f-  px  +  q  clearly  has  no 
factors  not  containing  x. 

Hence  x2  +  pas  -j-  q  =  (x  —  a)  (x  —  b)     .     .     .   (3) 

Performing  the  multiplication  in  (3),  we  have 
x2  +  2XV  +  Q  =  a'2  —  (ft  +  b)x  -\-  ab. 
Hence  we  have      a  +  b  =  — p  \  ,.-. 


ab  =       q. 

That  is,  in  a  quadratic  equation  where  the  coefficient  of  x*  is 
unity  and  all  the  terms  are  in  the  first  member,  the  swot  of  the 
roots  is  equal  to  the  coefficient  of  x  with  its  sign  changed,  and 
the  product  of  the  roots  is  equal  to  the  third  term. 

Thus,  dividing  the  general  equation  by  a,  it  becomes 

x2  +  hx  +''  =  <> (<r>) 

a         a 


TO  FORM  A   QUADRATIC  WHEN  THE  ROOTS  ARE  GIVEN.  293 

Adding  together  the  two  roots  of  (1),  Art.  137,  we  have 

x.  -J-  x„  =  ; 

a 

and  by  multiplication  we  have 

_  ^  _  (ba  _  4qc)  _  c 

4a-  a 

which  confirms  the  proposition. 

Hence  any  quadratic  may  be  expressed  in  the  form 

x1  —  (sum  of  roots)  x  +  product  of  roots  =  0   .    (6) 

By  this  principle  we  may  easily  form  a  quadratic  with 
given  roots.  Although  we  cannot  in  all  cases  find  the  roots 
of  a  given  equation,  it  is  easy  to  solve  the  converse  problem, 
namely,  the  problem  of  finding  an  equation  which  has  given 
rooW. 

These  relations  are  useful  in  verifying  the  solution  of  a 
quadratic  equation.  If  the  roots  obtained  do  not  satisfy 
these  relations,  we  know  there  is  some  error  in  the  work. 

Relations  analogous  to  those  above  hold  good  for  equations 
of  the  third  and  of  higher  degrees.  But  we  defer  the  proof 
to  a  subsequent  chapter. 

When  we  know  one  root  of  an  equation,  we  may  by 
division  lower  the  degree  of  the  equation.  Thus  if  a  is  one 
root  of  an  equation,  we  may  divide  it  by  the  factor  x  —  a. 

Note  1.  —  In  any  equation  the  term  which  does  not  contain  the 
unknown  quantity  is  frequently  called  the  absolute  term. 

A  quadratic  equation  cannot  have  more  than  two  roots.  For 
no  other  value  of  x  besides  a  and  b  can  make  (x  —  a)(x  —  b) 
in  (3)  equal  to  0,  since  the  product  of  two  factors  cannot 
equal  0  if  neither  factor  is  equal  to  0. 

It  may  therefore  be  inferred  that  a  cubic  equation  has  three 
and  only  three  roots  ;  and  that  in  any  equation  the  number  of 
roots  is  equal  to  the  degree  of  the  equation. 

Note  2.  —  The  student  must  carefully  distinguish  between  a  quad- 
ratic equation  and  a  quadratic  expression.  Thus,  in  the  quadratic 
equation  x'1  +  px  -f  q  =  0,  we  must  suppose  x  to  have  one  of  two 


294  EQUATIONS  HAVING   IMAGINARY  ROOTS. 

definite  values,  or  roots,  but  when  we  speak  of  the  quadratic  expres- 
sion x2  +  px  +  q,  without  saying  that  it  is  to  he  equal  to  zero,  we  may 
suppose  x  to  have  any  value  we  please. 

EXAMPLES. 

Form  the  quadratic  equation  whose  roots  are 

1.  2  aud  8.  Ans.  x2  —  5x  +  6  =  0. 

2.  3  and  -2.  a?  -  x  -  6  =  0. 
&  2  +  V.3  and  2  -  \]'d.                           x2  -  4x  +1  =  0. 

4.  6  and  8.  x?  -  Ux  +  48  =  0. 

5.  4  and  5.  x2  —  9x  +  20  =  0. 
141.  Equations  Having  Imaginary  Roots.  —  It  was 

shown  (Art.  137)  that  when  b2  is  less  than  4ac,  i.e.,  when 

b2  .  c 

——  is  less  than  -,  the  two  roots  are  imaginary.  Hence,  from 
4a2  a  * 

(5)  and  (G)  of  Art.   140,  the  roots  are  imaginary  when  the 

square  of  half  their  sum  is  less  than  their  product.     Now  it 

is  impossible  to  have  two  numbers  such  that  the  square  of 

half  their  sum  is  less  than   their  product,   which  may  be 

shown  as  follows : 

Let  a  represent  any  number  ;  and  suppose  it  to  be  divided 

into  twro  parts  — f-  x  and  "  —  x.     Then  the  product  is 

a2         o 

which  is  evidently  the  greatest  when  x2  is  the   least.     But 

when  a;'2  or  x  =  0  the  parts  are  each  -  ;  thus  the  product  of 

two  unequal  numbers  is  less  than  the  square  of  half  their 
sum.     Hence, 

The  square  of  half  the  sum  of  two  numbers  can  never  be 
less  than  their  product. 

If  then  any  problem  furnishes  an  equation  of  the  general 
quadratic  form  x1  +  px  +  7  =  0,  in  which  q  is  positive  and 

greater  than  the  square  of  ^-,  we  infer  that  the  conditions  of 


EQUATIONS  OF  HIGHER  DEGREE  THAN  THE  SECOND.    295 

the  problem  are  incompatible  with  each  other,  and  hence  the 
problem  is  impossible.     Thus, 

Let  it  be  required  to  divide  6  into  two' parts  whose  product 
shall  be  10. 

Let  x  =  one  of  the  parts, 
then  6  —  x     —  the  other. 

.-.     x(6  -  x)  =  10, 
whence  x  =  3  ±  V  —  1 . 

Thus,  the  roots  are  imaginary.  Now  we  know  from  the 
preceding  proposition,  that  the  number  6  cannot  be  divided 
into  any  two  parts  whose  product  will  be  greater  than  9. 
Hence  when  we  are  required  to  divide  6  into  two  parts  whose 
product  is  10,  we  are  required  to  solve  an  impossible  problem. 
Thus,  the  imaginary  root  shows  that  the  problem  is  impossible. 

142.  Equations  of  Higher  Degree  than  the  Second. 

—  There  are  many  equations  which  are  not  really  quadratic, 
but  which  may  be  reduced  to  the  quadratic  form,  and  solved 
by  the  methods  explained  in  this  Chapter.  An  equation  is 
in  the  quadratic  form  when  the  unknown  quantity  is  found  in 
two  terms,  and  its  exponent  in  one  term  is  twice  as  great  as 
in  the  other.     Thus, 

x*  -  9x2  =  -20,  (x2  +  x)2  +  4(.«2  +  x)  =  12, 
ax2"  +  bxn  +  c  =  0,  etc., 

are  in  the  quadratic  form,  and  may  be  solved  by  either  of 
the  preceding  rules  ;  care  however  should  be  taken  to  use 
the  one  best  adapted  to  the  example  considered. 

1.  Solve  xi  -  9x2  =  -20. 

Here  we  may  complete  the  square  and  solve  by  Rule  I, 
Art.  136,  or  we  may  write  out  the  value  of  x2  at  once  by 
Rule  II,  Art.  136  as  follows : 


x2  =  f  ±  V-  20  +  -V-  = 

=  f  ±  i  =  5,  or  4. 
x  =  ±VE,  or  ±2. 

Thus  there  are  four  roots,  ±sJ-'>,  ±2. 


296   EQUATIONS  OF  HIGHER  DEGREE  THAN  THE  SECOND. 

Otherwise  thus. 

Transposing  and  factoring  the  first  member, 
(a?  -  5)(jb3  -  4)  =  0. 

.-.     as2  —  5  =  0,  giving  x-  =   ±05, 
or  a;2  —  4  =  0,  giving  x  —  ±2. 

2.  Solve  oas2"  +  6a;"  +  c  =  0. 
Transpose  and  divide  by  a, 

B*  +   ^   =    _C. 


Art.  136,  Rule  II,  a-  =  -A  ±v/_c  +  _&1  =  V_-4ac 

2a      V      a       4a2  4a2 


-  ~b  ±  ^  -  4ac 
2a 
from  which  x  may  be  found  by  taking  the  n,h  root  of  both 
members. 

Note  1.  —  If  the  student  prefers,  he  may  let  xn  =  y;  then  x"M  =  y'-. 
Substituting,  the  equation  becomes 

«>J2  +  by  +  c  =  0. 
After  solving  for  y,  he  may  replace  tbe  value  of  y. 

3.  Solve   a:6  -  Gxs  =  16. 

Art.  136,  Rule  II,  x3  =  3  ±V/25  =  3  ±  5  =  8,  or  -1 
.-.     x  =  2,  or  -V2. 

4.  Solve  a?- *  +  -c-4  =  6. 

Solving  fora;-*,  a.--'  =  —  $  x  V^j  +  *  =  2,  or  -3. 
.-.     a;"1  =  16,  or  81. 
•'•      x       =  TS>  or  ST- 

5.  Solve  y/x*  +  12  +  VaJ2  +  12  =  6. 
Solving  for  Vxr  -j-  12, 


t/x*  +  12  =  _i   ±  yV,  +  |  =  — i  ±  §  =  2,  or  -3. 

ar  +  12  =  16,  or  .Si. 

a?  =  I,  or  69. 

x  =  ±2,  or  ±V69. 


EQUATIONS  OF  HIGHER  DEGREE  Til  AS  THE  SECOND.   297 


6.  Solve  x  +  \bx  +  10  =  8. 
By  transposing  a;,  and  squaring, 

5a  +  10  =  64  -  16a;  +  x2. 
.-.     x2  -  2\x  =  -54. 
Solving,  we  get  x  =  -2/-  ±  x£  =  18,  or  3. 

If  we  proceed  to  verify  these  values  of  x  by  substituting 
them  in  the  given  equation,  we  shall  find  that  3  satisfies 
the  equation,  but  that  18  does  not,  while  it  does  satisfy  the 
equation 

x  -  \jbx  +  10  =  8. 
Now  the  reason  is  this  :  the  equation  x2  —  21a;  =  —54, 
which  we  obtained  from  the  given  equation  by  transposing 
and  squaring,  might  have  been  obtained  as  well  from 
x  —  \Jbx  +  10  =  8,  since  the  square  root  of  a  quantity 
may  have  either  the  sign  +  or  —  prefixed  to  it;  i.e.,  the 
resulting  equation  x2  —  21a;  =  —54,  of  which  18  and  3  are 
the  roots,  would  be  obtained,  whether  the  sign  of  the  radical 
be  —  or  — .  Hence  we  see  that  when  an  equation  has  been 
reduced  to  a  rational,  form  by  squaring,  we  cannot  be  certain 
without  trial  whether  the  values  which  are  finally  obtained 
for  the  unknown  quantity  are  roots  of  the  given  equation. 

7.  Solve  x2  -  7x  +  sjx2  -  7a;  +  18  =  24. 

Acid  18  to  both  members  in  order  that  the  equation  may  be 

in  the  quadratic  form. 

x*  _  7aj  +  18  +  sjx2  -  7x  +  18  =  42. 


Solving,  Sfx1  -7a,-+18=-f±V/42  +  i 

=  —  \  ±  -1/  =  6,  or  -7 
.-.     x2  —  7x  +  18  =  36,  or  49. 
Solving  the  first  quadratic,  we  obtain      x  —  9,  or  —2. 
Solving  the  second  quadratic,  we  obtain  x  =  \{1  ±  VI 73), 
Only  the  first  two  values  are  roots  of  the  given  equation 
the  other  two  are  roots  of  the  eq nation 

ar  -  7x  -  V'aJ2  —  1x  +  18  =  24. 


298  SOLUTIONS  BY   FACTORING. 

8.    Solve  x*  -  4x»  -  2x*  +  12as  -  16  =    0. 

We  proceed  to  form  a  perfect  trinomial  square  with  the 
first  two  terms  and  a  part  of  the  third.  The  square  root  of 
this  square  is  evidently  (x2  —  2a;),  the  square  of  which  is 
a4  —  Ax3  +  Ax2 ;  having  added  Ax2  to  the  equation  we  must 
now  subtract  Ax1.     Hence  the  equation  becomes 

x4  -  Ax3  +  Ax-  -  Gx2  4-  12a;  -  16  =    0, 
or  (a?  -  2x)2  -  G(x2  -  2x)  =  16, 

which  is  in  the  quadratic  form,  and  may  be  solved  as  those 
above. 

Hence  x2  -  2x  =  3  ±  ^10  +  9  =  8,  or  -2. 


.-.     x  =  1  ±  ^8  +  1,  or  1  ±  V-2  +  1. 
.-.     x  =  4,  or  -2,  or  1  ±  f^l. 
Solve  the  following : 
9.    a;4  -  13a?  +  36  =  0.  Ans.  ±2,  ±3. 


10.  x2  +  yV  +  9  =  21.  ±4. 

11.  9v/;f2  -  9x  +  28"  +  9a;  =  x2  +  36.  12,  -3. 

12.  x§  +  x%  =  1056.  64,  (-33)*. 

13.  (x2  -  9)2  -  11  (x2  —  2)  =  3.  ±5,  ±2. 

14.  a,-4  -  8xs  +  10.r  +  24a;  +  5  =  0.  5,  -1,  2  ±  fa 

143.  Solutions  by  Factoring.  —  By  the  principles  of 
Art.  139,  many  equations  of  a  higher  degree  than  the  second 
ma}'  be  solved,  which  cannot  be  reduced  to  the  quadratic 
form.     If  an  equation  can  be  reduced  to  the  form 

(x  -  a)X  =  0, 
in  which  X  represents  an  expression  involving  aj,  we  have 
either  «-a-0,    orX=0; 

therefore  x  =  a,  is  one  value  of  x  ;  and  if  we  solve  the 
equation  X  =  0,  we  shall  have  the  other  values  of  x.  Hence 
whenever  we  have  one  factor  of  an  equation,  we  have  at 
Least  one  root,  and  by  division  we  may  lower  the  degree  of 
the  equation  by  one  (Art.  1 10).     Thus 


SOLUTIONS  BY  FACTORING.  2dd 

1.  Solve  (x  -  5)  (x2  -  3a;  +  2)  =  0. 

Here  the  first  member  is  zero  either  when  x  —  5  =  0,  or 
when  x2  —  ox  +  2  =  0  ;  and  in  no  other  case.  Hence  we 
have 

x  —  5  =  0,  or  x1  -  Sx  +  2  =  0. 

From  the  first  we  have  x  =  5  ;  and  the  other  roots  of  the 
equation  are  those  given  by 

x"  -  3x  +  2  =  0, 
that  is,  (x  -  2)  (a;  -  1)  =  0. 

Thus  the  cubic  equation  (as  —  5)  (x2  —  3x  +  2)  =  0  has 
the  three  roots  5,  2,  and  1. 

The  difficulty  to  be  overcome  in  this  method  consists  in 
resolving  the  equation  iuto  factors  ;  and  facility  in  separat- 
ing expressions  into  factors  can  be  acquired  only  by  experi- 
ence. 

2.  Solve  xs  -  1  =  0. 

Since  x3  -  1  =  (a;  -  l)(x2  +  x  +  1), 

we  have  (x  -  l)(x*  +  x  +  1)  =  0. 

.-.     x  —  1  =  0,  or  ar  +  a;  +  1  =  0, 

the  roots  of  which  are  1,  or  —  \  ±  V— f. 

Hence  there  are  three  roots  of  the  equation  x3  =  1,  one 
being  real  and  the  other  two  imaginary.  Thus  there  are 
three  numbers  whose  cubes  are  equal  to  1  ;  that  is,  there 
are  three  cube  roots  of  1. 

2 


3.    S( 

jive   x  —  1  =  2  +  -p. 
yjx 

Ans.  4. 

4.       ' 

:'      2x3  -  a;2  —  6a;  =  0. 

0,  2,  -f. 

5.       ■ 

"      x3  -\-  x2  —  4x  =  4. 

-1,  2,  -2. 

6. 

"      a3  -  3x  =  2. 

-1,  2. 

7. 

«       a;2  -A  =  If.     ' 
3x          9 

2 

3' 

Hi  ±  Vio). 

300      PROBLEMS    LEADING    TO    QUADRATIC   EQUATIONS. 

144.  Problems  Leading  to  Quadratic  Equations 
of  One  Unknown  Quantity. —  We  shall  now  give  some 
examples  of  problems  wliieh  lead  to  pure  or  affected  quad- 
ratic equations  of  one  unknown  quantity. 

In  the  solution  of  such  problems,  the  equations  are  found 
on  the  same  principles  as  in  problems  producing  simple 
equations  (Art.  61). 

EXAMPLES. 

1.  Find  two  numbers  such  that  their  sum  is  15,  and  their 
product  is  54. 

Let  x  =  one  of  the  numbers, 

then  15  —  x  =  the  other  number. 

Hence  from  the  conditions,  we  have 
a(15  —  x)  =  54, 
j»r  x"  —  15a;  =  —54. 

Solving,  we  get  x  =  9,  or  6. 

If  we  take  x  =  9  we  have  15  —  x  =  6  ;  and  if  we  take 
x  =  6  we  have  15  —  x  =  9.  Thus,  whichever  value  of  x 
we  take,  we  get  for  the  two  numbers  G  and  9.  Hence, 
although'  the  equation  gives  two  values  of  x,  yet  there  is 
really  only  one  solution  of  the  problem. 

2.  A  man  buys  a  horse  which  he  sells  agaiu  for  $96  ;  he 
finds  that  he  thus  loses  one-fourth  as  much  per  cent  as  the 
horse  cost  him  :  find  the  price  of  the  horse. 

Let    x  =  the  price  of  the  horse  in  dollars, 

x2 

then  —  =  the  man's  loss  in  dollars. 
100 

Hence  from  the  conditions,  we  have 


400 


9G. 


Solving,  we  get  x  =  240  or  1G0. 

That   is,  the   price  was   either,  $210  or  $1G0,    for  each   of 
these  values  satisfies  the  conditions  of  the  problem. 


EXAMPLES.  301 

3.  A  train  travels  300  miles  at  a  uniform  rate  ;  if  the  rate 
had  been  5  miles  an  hour  more,  the  journey  would  have  taken 
two  hours  less  :  find  the  rate  of  the  train. 

Let  x  =  the  rate  the  train  runs  in   miles  per 

hour  ; 
then  300  -J-  x  =  the  time  of  running  on  the  first  sup- 

position ; 
and  300  -f-  (re  -f-  5)  =  the   time  of  running  on  the  second 
supposition. 
300     =  300      2 
x  -f  5         x 
Solving,  we  get  x  =  25,  or  —30. 

Only  the  positive  value  of  x  is  admissible,  and  thus  the 
train  runs  25  miles  per  hour. 

Note.  —  In  the  solutions  of  problems  it  often  happens  that  the 
roots  of  the  equation,  which  is  the  Algebraic  statement  of  the  relation 
between  the  magnitudes  of  the  known  and  unknown  quantities,  do 
not  all  satisfy  the  conditions  of  the  problem.  The  reason  of  this  is 
that  the  Algebraic  statement  is  more  general  than  ordinary  language; 
and  the  equation,  which  is  a  proper  representation  of  the  conditions, 
will  also  express  other  conditions.  Thus,  the  roots  of  the  equation 
are  the  numbers,  whether  positive,  negative,  integral,  or  fractional, 
which  satisfy  that  equation  ;  but  in  the  problem  there  may  be  restric- 
tions on  the  numbers,  expressed  or  implied,  which  cannot  be  retained 
in  the  equation.  If  for  instance,  one  of  the  roots  of  an  equation  is  a 
fraction,  it  cannot  be  a  solution  of  a  problem  which  refers  to  a  number 
of  men,  for  such  a  number  must  be  integral.     Thus 

4.  Eleven  times  the  number  of  men  in  a  group  is  greater 
by  twelve  than  twice  the  square  of  the  number :  find  the 
number  of  men  in  the  group. 

Let  x  =  the  number  of  men  ;  then  we  have 
llaj  =  2x-  +  12, 
or  2a;2  -  11a;  =  -12. 

Solving,  we  get  x  =  4,  or  1^. 

Thus,  there  are  4  men  ;  the  value  1|  is  plainly  inadmissi- 
ble. 


302  EXAMPLES. 

5.  Eleven  times  the  number  of  feet  in  the  length  of  a  rod 
is  greater  by  twelve  than  twice  the  square  of  the  number  of 
feet:  how  long  is  the  rod? 

This  question  leads  to  the  same  equation  as  Ex.  4,  only 
here  we  cannot  reject  the  fractional  result,  since  the  rod  may 
be  either  4  feet  long  or  1^  feet  long. 

6.  The  square  of  the  number  of  dollars  a  man  possesses 
is  greater  by  600  than  ten  times  the  number :  how  much  has 
the  man? 

Let  x  =  the  number  of  dollars  the  man  has. 
Then  x1  =  10a?  +  GOO. 

Solving,  we  get  x   =  30,  or  —20. 

Both  these  values  are  admissible,  since  a  negative  posses- 
sion is  a  debt  (Art.  20) . 

7.  The  sum  of  the  ages  of  a  father  and  his  son  is  80  years  ; 
also  one-fourth  of  the  product  of  their  ages,  in  years, 
exceeds  the  father's  age  by  240  :  how  old  are  they? 

Lqt  x  =  the  father's  age  in  years  ; 

then  80  —  x  =  the  son's  age  in  years. 

Hence  Ja?(80  -  x)  =  x  +  240, 
or  x2  —  IGx  =  —960. 

.-.     x  =  60,  or  16. 

Thus  the  father  is  60  and  the  son  20  years  old. 

The  second  solution  16  is  not  admissible,  since  it  would 
make  the  father  younger  than  his  son. 

Note.  —  The  student  should  examine  each  root  of  every  equation 
to  see  if  it  satisfies  the  conditions  of  the  problem,  and  reject  those 
which  do  not. 


5.  A  cistern  can  be  filled  by  two  pipes  in  33^-  minutes  ;  if 
the  larger  pipe  takes  15  minutes  less  than  the  smaller  to  fill 
the  cistern,  find  in  what  time  it  will  be  filled  by  each  pipe 
singly.  Ans.  75  and  60  minutes. 

9.  A  person  selling  a  horse  for  $72,  finds  that  his  loss  per 
cent  is  one-eighth  of  the  number  of  dollars  that  he  paid  for 
the  horse  :  what  was  the  cost  price?  Ans.  $80,  or  $720. 


EXAMPLES   OF    QUADRATICS.  303 

10.  Divide  the  number  10  into  two  parts  such  that  their 
product  added  to  the  sum  of  their  squares  may  make  76. 

Ans.  4,  6. 

11.  Find  the  number  which  added  to  its  square  root  will 
make  210.  Ans.  196. 

/T2?  A  and  B  together  can  do  a  piece  of  work  in  14f  days  ; 
and  A  alone  can  do  it  in  12  days  less  than  B  alone  :*  find  the 
time  in  which  A  alone  can  do  the  work.  Ans.  24  days. 

13.  A  company  dining  together  at  an  inn,  find  their  bill 
amounts  to  $35  ;  two  of  them  were  not  allowed  to  pay,  and 
the  rest  found  that  their  shares  amounted  to  $2  a  man  more 
than  if  all  had  paid  :  find  the  number  of  men  in  the  company. 

Ans.  7. 

14.  The  side  of  a  square  is  110  inches  long  :  find  the  length, 
and  breadth  of  a  rectangle  which  shall  have  its  perimeter 
4  inches  longer  than  that  of  the  square,  and  its  area  4  square 
inches  less  than  that  of  the  square.  Ans.  126,  96. 

Note.  — We  will  conclude  this  chapter  with  the  following  examples. 
In  solving  them  care  must  he  taken  to  select  the  method  best  adapted 
to  the  example  considered.  Many  of  them  may  be  solved  by  special 
methods  (Arts.  137,  138,  139);  but  the  methods  of  Art.  136  are  the 
most  important,  and  will  solve  every  example. 

EXAMPLES    OF    QUADRATICS. 

Ans.  ±h. 

±2. 

±y/3. 

2,  3. 

1,  9. 

\,  -13. 


1. 

0:^6  +  a;2  =  1  +  x2. 

2. 

lx2  +  8         x2  +  4         x2 
21-           8x2  -  11        3 
o 

n 

•>. 

«  +  vi  +  ^-=v/i  +  a;3- 

4. 

1         ,         i 

x  +  ^2  —  ar       «  —  \/2  —  x2 

_  X 

~  2 

5. 

x2  —  ox  =  —6. 

6. 

x2  —  10a;  =  —9. 

7. 

x2  +  12a;  =  13. 

304  EXAMPLES. 

8.  x-  +  9a  =  22.  Am.  2,  -11. 

9.  x2  +  2x  =  143.  11,  -13. 


10.    15.x2  -  2a.r  =  a2. 


11. 

21x2  =  2ax  +  3a2 

12. 

x  +  3        2x  -  1 
2a  -  7        x  -  3 

13. 

1       _       1 
1+05         3  —  a 

11. 

4                5 

=  &. 


a  +  1             x  —  1 

19. 

a;  -f  1       a  —  2  _  „ 
a  -  1       a  +  2  ~  *' 

20. 

1                 2       _  3 

a;  -  2       a;  -f  2        &' 

21. 

3                       1 

2(ar  -  1)        4(a;  +  1) 

22. 

5            3  _       14 

x  +  2       a;        a?  +  4 

23. 

4                5                12 
a:  +  1        x  +  2        a;  +  3 

24. 

a;  +  1        a;  -  1  _  2x  -  1 

x  +  2 


a     _a 
3'        5" 

**  ~3 


M- 


3,  -£• 


x  —  1        a;  +  2       a) 

15.  (a;  +  l)(2a;  +  3)  =  4a;2  -  22.  -  5,  -2£. 

16.  2  +  ^  -  ^LzUf  =  l  _  aj  +  x2.  1,  2. 


17.    8s  +  11  +  -  =  — •  7,  -£,. 

a;  7 

18<    3(s  -  1)  _  2(a-  +  1)  =  5  f  _3 


3,  -I- 
3,  -4f. 

3,  -5. 

3,  -If 

3,  -If 

4,0. 


j  +  2       a  —  S        \x-Sj 


EXAMPLES. 

305 

26. 

x  —  1        x  -  2  _  2x  +  13 
•Hi       x  +  2  ~"  a;  +  16  ' 

^4ns.  5,  —  1^. 

27. 

a;  +  1       as  +  2  _  2a?  4-  13 

a;  —  1       sc  —  2         a;  +  1 

5,  1*. 

28. 

2x  -  1       Sx  —  1  _  5a;  -  11 

a:  +  1     '     a;  +  2          a:—  1 

5,  -1£. 

29. 

14a;  _9       a;2  -  3 
8a;  —  3         x  +  1 

■    1        1,1 

a;  +  -        1  +  - 

2f,0. 

30. 

—=  +  —?=* 

a; 1 

a;               a; 

3,  -If- 

31. 

V/Ia;  +  2       4  -  y/a; 
4  +  V^85             V^ 

4. 

32. 

a2ar  -  2a3x  +  a4  -  1  =  0. 

a 

33. 

4a2a;  =  (a2  -  b2  +  a;)2. 

(a  ±  o)2. 

34. 

a;    .    a  _  a;   ,    & 

a       x       b       x 

±V^&. 

35. 

(3a2  +  62)(x-2  -  x  +  1)  =  (362  + 

a8)  (»2  +  aJ+  1). 

Ans. 


a  —  b     a  +  b 

36. 

1                   1     , 

o       a; 

—  a,  —b. 

a  +  b  +  x       a 

37 

a  +  c(a  +  x)       a 
a  +  c(a  —  x) 

+  a;            a 

a,  °(1+c). 
c(2c  +  3) 

a;           a  —  2ca; 

»  +  2  _  I  =  o. 
a       a;       a 

38. 

1  ±  y/l  -a2- 

39 

*                   (7/1         7) 

*N~  —                          1 

.    a,  —b. 

ya  +  V^& 


(o62)  -2  +  (a86)  "i 


306  EXAMPLES. 

Form  the  equation  whose  roots  are 

40.  1  ±  5.  Ans.  x2  —  2x  -24  =  0. 

41.  -$,  f.  35a;2  +  13a:  -  12  =  0. 

42-  £ir|  -frrf    (^2  ~  *>*  +  4^  -  Pa  +  ff3  =  o. 

43.  7  ±  205.  a2  -  14a>  4-  29  =  0. 

44.  ±2y/3  -  5.  x2  +  10a:  +  13  =  0. 

45.  -p  ±  2y/2g.  a;2  +  2pa;  +  p2  -  8g  =  0. 

Show  that  the  roots  of  the  following  equations  are  real : 

46.  x2  -  2ax  +  a?  -  b2  -  c2  =  0. 

47.  (a  -  6  +  c)x2  +  4 (a  -  &)a  +  (a  -  6  -  c)  =  0. 

For  what  values  of  m  will  the  following  equations  have 
equal  roots? 

48.  a;2  -  15  -  m(2x  -  8)  =  0.  Ans.  3,  5 

49.  a;2  -  2sc(l  +  3m)  +  7(3  +  2m)  =  0.  2,  — V0"- 

EXAMPLES    OF    EQUATIONS     REDUCIBLE    TO 
QUADRATICS. 

50.  a;4  -  5a;2  +  4  =  0.  Ans.  ±1,  ±2. 

51.  (a?2  -  a;)2  -  S(x2  -  x)  +  12  =  0.  3,  -1,  ±2. 

_1    4-  i/_'>7 

52.  (a;2  +  a?)(a*»  +  a;  +  1)  =  42.    2,  -3, g  — 

53.  (a;  +  ^Y+  4^  +  ~\  =  12.  i,  _3±  2y/2. 


54. 
55. 

x2  +  3  -  \/2x2  -  3a;  +  2  =  §(a>  +  1).               1,  J. 
35*  _|_  2a;3  -  liar  +  4.«  +  4  =  0.                             1,  2. 

56. 

57. 

.  ,                                           «    j_  aV2\/2  -  1 
x*  +  4a8a;  =  a4.                            — F  ±  *t= • 

y/2             y/2 

1+.^    _  , 

(1  +  »)< 

4«8.  1  4-  ^3  ±  ^3  +  2^3.  1  -  V3  ±  ^3  -  2\<3. 


EXAMPLES.  307 

58.    27a;2  -  ~  +  y  =  —  -  — „  -f  5. 

3a;2         3  3a;        3a;2 

Ans.  2,  -If,  A(-2  ±  V/-2GG). 
Solve  as  explained  in  Art.  143  : 

Ans.  1,  |-(-l  ±  v/^7) 

3,  k-3  ±  y/£3) 

hH-1  ±  \/-35) 


Find  a  number  whose  square  diminished   by  119    is 
equal  to  ten  times  the  excess  of  the  number  over  8.  Ans.  13. 

67.  A  man  is  five  times  as  old  as  his  son,  and  the  sum  of 
the  squares  of  their  ages  is  equal  to  210G  :  find  their  ages. 

Ans.  45,  9. 

68.  If  a  train  traveled  5  miles  an  hour  faster  it  would 
take  one  hour  less  to  travel  210  miles:  what  time  does  it 
take?  Ans.  7  hours. 

69.  The  sum  of  the  reciprocals  of  two  consecutive  numbers 
is  £f  :  find  them.  Ans.  7,  8. 

70.  The  perimeter  of  a  rectangular  field  is  500  yards,  and 
its  area  is  14400  square  yards  :  find  the  length  of  the  sides. 

Ans.  90  and  160  yards. 

71.  A  number  of  two  digits  is  equal  to  twice  the  product 
of  the  digits,  and  the  digit  in  the  ten's  place  is  less  by  3 
than  the  digit  in  the  unit's  place  :  what  is  the  number? 

Ans.  36. 

72.  The  sum  of  a  certain  number  and  its  square  root  is 
42  :  what  is  the  number?  Ans.  36. 


308  EXAMPLES. 

73.  A  rectangular  court  is  ten  yards  longer  than  it  i3 
broad;  its  area  is  1131  square  yards:  find  its  length  and 
breadth.  Ans.  39  and  29. 

74.  One  hundred  and  ten  bushels  of  coal  were  divided 
among  a  certain  number  of  persons ;  if  each  person  had 
received  one  bushel  more  he  would  have  received  as  many 
bushels  as  there  were  persons :  find  the  number  of  persons. 

Ans.  11. 

75.  A  cistern  can  be  supplied  with  water  by  two  pipes  ; 
by  one  of  them  it  would  be  filled  G  hours  sooner  than  by  the 
other,  and  by  both  together  in  4  hours :  find  the  time  in 
which  each  pipe  alone  would  fill  it.  Ans.  G,  12. 

(76J  Two  messengers  A  and  B  were  sent  at  the  same  time 
to  a  place  90  miles  distant ;  the  former  by  riding  one  mile 
per  hour  more  than  the  latter  arrived  at  the  end  of  his 
journey  one  hour  before  him :  find  at  what  rate  per  hour 
each  traveled.  Ans.  10,  9  miles. 

77.  A  person  rents  a  certain  number  of  acres  of  land  for 
$280  ;  he  keeps  8  acres  in  his  own  possession,  and  sublets 
the  remainder  at  $1  per  acre  more  than  he  gave,  and  thus  he 
covers  his  rent  and  has  $8  over :  find  the  number  of  acres. 

Ans.  56. 

78.  From  two  places  320  miles  apart,  two  persons  A  and 
B  set  out  in  order  to  meet  each  other.  A  traveled  8  miles  a 
day  more  than  B  ;  and  the  number  of  days  in  which  they 
met  was  equal  to  half  the  number  of  miles  B  went  in  a  day  : 
find  how  far  each  traveled  before  they  met.     Ans.  192,  128. 

79.  A  certain  number  is  formed  by  the  product  of  three  con- 
secutive numbers,  and  if  it  be  divided  by  each  of  them  in  turn, 
the  sum  of  the  quotients  is  -17  :  find  the  number.       Ans.  GO. 

■so.  A  boat's  crew  row  3  J  miles  down  a  river  ami  back 
again  in  1  hour  and  40  minutes  :  supposing  the  river  to  have 
a  cm-rent  of  2  miles  per  hour,  find  the  rate  at  which  the  crew 
would  row  in  still  water.  Ans.  5  miles  per  hour. 

.si.  A  person  rents  :i  certain  number  of  acres  of  land 
for$33G  ;  he  cultivates  1  acres  himself,  and  letting  the  rest  for 


EXAMPLES.  309 

$2  an  acre  more  than  he  pays  for  it,  receives  for  this  portion 
the  whole  rent,  $336  :  find  the  number  of  acres.  Ans.  28  acres. 

82.  A  person  bought  a  certain  number  of  sheep  for  $140  ; 
after  losing  two  of  them  he  sold  the  rest  at  $2  a  head  more 
than  he  paid  for  them,  and  by  so  doing  gained  $4  by  the 
transaction  :  find  the  number  of  sheep  he  bought.     Ans.  14. 

83.  A  man  sends  a  lad  to  the  market  to  buy  12  cents' 
worth  of  oranges ;  the  lad  having  eaten  a  couple,  the  man 
pays  at  the  rate  of  one  cent  for  15  more  than  the  market 
price  :  how  many  did  the  man  get  for  his  12  cents?     Arts.  18. 

84.  What  are  eggs  a  dozen  when  two  more  in  a  shilling's 
worth  lowers  the  price  one  penny  per  dozen  ?  Ans.  Ninepence. 

85.  Two  men  were  employed  at  different  wages,  and  paid 
at  the  end  of  a  certain  time  ;  the  first  received  £4.  16s.,  and 
the  second,  who  had  worked  for  6  days  less,  received  £2.  14s. 
If  the  second  had  worked  all  the  time  and  the  first  had 
omitted  6  days,  they  would  have  received  the  same  sum : 
how  many  days  did  each  work,  and  what  were  the  wages  of 
each?  Ans.  24  days  at  4s.,  and  18  days  at  3s.  per  day. 

/86^)  A  man  bought  a  certain  quantity  of  meat  for  $2.16; 
if  meat  were  to  rise  in  price  one  cent  per  lb.  he  would  get  3 
lbs.  less  for  the  same  sum :  how  much  meat  did  he  buy  ? 

Ans.  27  lbs. 

87.  The  price  of  one  kind  of  sugar  per  stone  of  14  lbs.  is 
Is.  9d.  more  than  that  of  another  kind;  and  8  lbs.  less  of 
the  first  kind  can  be  got  for  £1  than  of  the  second  :  find  the 
price  of  each  kind  per  stone.  Ans.  8s.  9d.,  7s. 

88.  A  person  spent  a  certain  sum  of  money  in  goods, 
which  he  sold  again  for  £24,  and  gained  as  much  per  cent  as 
the  goods  cost  him  :  find  what  the  goods  cost.         Ans.  £20. 

89.  A  person  drew  a  quantity  of  wine  from  a  full  vessel 
which  held  81  gallons,  and  then  filled  up  the  vessel  with 
water ;  he  then  drew  from  the  mixture  as  much  as  he  before 
drew  of  pure  wine ;  and  it  was  found  that  64  gallons  of  pure 
wine  remained :   find  how  much  he  drew  each  time. 

Aits.  9  gallons. 


310  EXAMPLES. 

90.  A  certain  company  of  soldiers  can  be  formed  into  a 
solid  square ;  a  battalion  consisting  of  seven  such  equal 
companies  can  be  formed  into  a  hollow  square,  the  nun 
being  four  deep.  The  hollow  square  formed  by  the  battalion 
is  sixteen  times  as  large  as  the  solid  square  formed  by  one 
company  :  find  the  number  of  men  in  the  compan}-.    Ans.  64. 

91.  Find  that  number  whose  square  added  to  its  cube  is 
nine  times  the  next  higher  number.  Ans.  3. 

92.  A  courier  proceeds  from  one  place  P  to  another  place 
Q  in  14  hours  ;  a  second  courier  starts  at  the  same  time  as 
the  first  from  a  place  10  miles  behind  P,  and  arrives  at  Q 
at  the  same  time  as  the  first  courier.  The  second  courier 
finds  that  he  takes  half  an  hour  less  than  the  first  to  go  20 
miles  :  find  the  distance  from  P  to  Q.  Ans.  70  mile.^. 

93.  A  vessel  can  be  filled  with  water  by  two  pipes ;  by 
one  of  these  pipes  alone  the  vessel  would  be  filled  2  hours 
sooner  than  by  the  other;  also  the  vessel  can  be  filled  by 
both  pipes  together  in  1|  hours:  find  the  time  which  each 
pipe  alone  would  take  to  fill  the  vessel.     Ans.  5  and  3  hours. 

94.  There  are  three  equal  vessels,  A,  B,  and  C  ;  the  first 
contains  water,  the  second  brandy,  and  the  third  brandy  and 
water.  If  the  contents  of  B  and  C  be  put  together,  it  is 
found  that  the  fraction  obtained  b}-  dividing  the  quantity  of 
brandy  by  the  quantity  of  water  is  nine  times  as  great  as  if 
the  contents  of  A  and  C  had  been  treated  in  like  manner : 
find  the  proportion  of  brandy  to  water  in  the  vessel  C. 

Ans.  Equal. 

95.  A  person  lends  $5000  at  a  certain  rate  of  interest ; 
at  the  end  of  a  year  he  receives  his  interest,  spends  $25  of 
it,  and  adds  the  remainder  to  his  capital ;  he  then  lends  his 
capital  at  the  same  rate  of  interest  as  before,  and  at  the  end 
of  another  }Tear  finds  that  he  has  altogether  85382  :  find  the 
rate  of  interest.  Ans.  1  per  cent. 


SIMULTANEOUS    QUADRATIC  EQUATIONS.  311 


CHAPTER    XV. 

SIMULTANEOUS     QUADRATIC     EQUATIONS. 

145.  Simultaneous  Quadratic  Equations.— We  shall 
now  consider  some  of  the  most  useful  methods  of  solving 
simultaneous  equations,  one  or  more  of  which  may  be  of  a 
degree  higher  than  the  first.  It  should  be  remarked  that  it 
is,  in  general,  impossible  to  solve  a  pair  of  simultaneous 
quadratic  equations  ;  for,  if  we  eliminate  one  of  the  unknown 
quantities,  the  resulting  equation  will  be  of  the  fourth  degree 
with  respect  to  the  other  unknown  quantity,  and  we  cannot,  in 
general,  solve  an  equation  of  a  higher  degree  than  the  second. 

There  are  several  cases  however  in  which  a  solution  of 
two  equations  may  be  effected,  when  one  or  both  of  them 
are  of  the  second  or  some  higher  degree.  Various  artifices 
are  employed  for  the  solution  of  such  equations,  the  proper 
application  of  which  must  be  learned  by  experience. 

146.  Case  I.  When  One  of  the  Equations  is  of 
the  First  Degree.  —  This  case  may  be  solved  by  the 
following 

Rule.  From  the  simple  equation  find  the  value  of  one  of 
the  unknown  quantities  in  terms  of  the  other,  and  substitute 
this  value  in  the  other  equation. 

1.    Solve  Zx  +  4?y  =  18 (1) 

5x2  -  3xy  =    2 (2) 

From  (1)  we  have  y  =  - — -  ; («*) 

and  substituting  in  (2), 

5a°  -  3»<18  ~  3-c>  =  2. 
4 

.-.     20a8  -  54.x-  +  9a:2  =  8, 

or  29a;2  -  54.x  =  8. 


312  SIMULTANEOUS    QUADRATIC   EQUATIONS. 


Solving,  we  get 

x  = 

2, 

or 

4     . 

and  by  substituting  in 

(3), 

y  = 

3, 

or 

267 

"58  '• 

Solve  the  following : 

2.   3a>  -  4ty  =  5, 

Ans, 

,  a;  =  3, 

or 

-H^ 

3X*  _  xy  _  stf 

=  21. 

y  =  i, 

or 

3.    ox   —    y  =  17, 

x  =  4, 

or 

8 

1j  ' 

SB?  =   12. 

y  =  3, 

or 

-  20. 

4.    2a;  -  5y  =    3, 

a?  =4, 

or 

-¥> 

x2  +  xy  =  20. 

2/=  1, 

or 

-ft- 

5.    Ax   —  by  =     1, 

a?  =  4, 

or 

-if, 

2x2  -  xy  +  3?/2  +  3aj  - 

-42/ 

17. 

2/  =  3, 

or 

_    4JL 

147.   Case  II.    Equations  of  the  Form  x  ±  y  =  a, 
and  xy  =  b ;  or  x-  +  y2  =  a,  and  jt/  =  6. 

1.  Solve  a?  +  y  ==     15.     .     .     .  (1) 

a#  =     36  .     .     .     .   (2) 

Square  (1),  x2  +  2xy  +  .7-  =  225;      .      .     .    (3) 

multiply  (2)  by  4,  4xy  =144;      .     .     .    (4) 

subtract  (4)  from  (3),  x2  —  2xy  +  y2  =     81 ; 
extract  the  square  root,  ®  —  y   =  ±9.       ...    (5) 

Combining  (0)  with  (1),  we  have  the  two  cases 
x  +  y  =  15,  ^     a;  +  y  =     15, 
x  -  y  =     9.j      x  -  y  =  -!). 
from  which  we  have  a;  =  12,  j  .r  =       3, 

2/  =    3. 1  ?/  =     12. 

2.  Solve  a-2  +  ?/2  =  25 (1) 

•'//  =12 (2) 

Multiply  (2)  by  2  ;  then  by  addition  and  subtraction  we  have 

x2  +  2xy  +  y2  =    49, 

x2  -  2.<v/  +  //-  =       1 . 

.-.     x  +  ?/  =   ±7, 

a;  —  y  =  ±1. 


SIMULTANEOUS    QUADRATIC   EQUATIONS.  313 

We  have  now  four  cases  to  consider,  namely, 
x  +  y=l)      x  +  y=      1\      aj  +  y=-7)      x  +  y=-7) 
x '  —  y  =  1 1     a?  —  y  =  —  1 J     x  —  y  =      1  j     »  —  y  =  - 1 ) 

From  which  the  values  of  as  are  4,  3,  —3,  —  4  ; 

and  the  corresponding  values  of  y  are  3,  4,  —4,  —3. 

Thus  there  are  four  pairs  of  values,  two  of  which  are  given 
by  x  —  ±4,  y  =  ±3,  and  the  other  two  by  x  =  ±3,  y  =  ±4, 
it  being  understood  that  in  both  cases  the  upper  signs  are  to 
be  taken  together,  and  the  lower  signs  are  to  be  taken 
together. 

These  are  the  simplest  forms  that  occur,  but  they  are 
specially  important,  since  the  solution  of  a  large  number  of 
other  forms  is  dependent  upon  them.  Our  object,  as  a  rule, 
is  to  solve  the  given  equations  symmetrically,  by  finding  the 
values  of  x  -f-  y  and  x  —  y,  which  we  can  always  do  as  soon 
as  we  have  obtained  the  product  of  the  unknown  quantities 
and  either  their  sum  or  their  difference. 

Any  pair  of  equations  of  the  form 

x2  ±  pxy  +  y2  =  a2 (1) 

x  ±  y  =  b (2) 

where  p  is  any  numerical  quantity,  can  be  reduced  to  one 
of  the  forms  above  considered ;  for  by  squaring  (2)  and 
combining  with  ( 1 ) ,  we  obtain  an  equation  to  find  xy ;  the 
solution  cau  then  be  completed  by  the  aid  of  equation  (2). 
Thus 

3.    Solve  x2  +  xy  +  y2  =  333    .     ._    .     .     .   (1) 

x-y  =  '6.     .     .'    .     .     .   (2) 

Square  (2),   ■     x2  -  2xy  +  y2  =  9 (3) 

Subtract  (3)  from  (1),  Sxy  =  324. 

.'.     xy  =  108 (4) 

Add  (1)  and  (4)  and  extract  square  root, 

x  +  y  =   ±21 •   (5) 

From  (2)  and  (5)  x  —  12,  or  —9. 

y  =  9,  or  —12. 


314  SIMULTANEOUS    QUADRATIC   EQUATION*. 

4.    Solve  I-I=i n> 

x       y 

1    +    1-5  ,.)\ 

Square  (1),  1  -  A  +  1   =  x (3) 

x-      xy      y 

Subtract  (3)  from  (2),      A  =  |       .......   (4) 

Add  (2)  and  (4)  and  take  the  square  root, 

-+-  =   ±1 (5) 

x       y 


From  (1)  and  (5),       -  =  f,  or  —\. 


1 

x 

1        i 

-  =  *,  or  — £. 

y  ^ 

.«.     x  =  |,  or  —3,  and  y  =  3,  or  — |. 
Solve  the  following  : 

5.  x  —  y  =12,  Ans.  x  =  17,      or  —   5, 

xy  =  85.  y  =     5,       or  —17. 

6.  x  +  ?/  =  74,  x  =  53,      or      21, 

xy  =1113.  y  =  21,      or      53. 

7.  a;  -f-  ?/  =  84,  x  =  71,       or       13, 

xy  =  923.  ?/  =13,      or      71. 

8.  of  +  if  =  1)7,  aj  =     i»,      or        4, 
a;    -f-  y   =13.  y  =,    4,      or        9. 

9.  a;  +  y  =  9,  a;  =     5,      or        4, 
x-  +  .ry  +  if  =  61.  ?/  =     I,      or        5. 

148.  Case  III.    When  the   Two  Equations   Con- 
tain a  Common  Algebraic  Factor. 

1  v i  le.    Divide  <>n<'  equation  by  the  other,  ami  caned  the 
cum  mint  factor. 


SIMULTANEOUS    QUADRATIC   EQUATIONS.  315 

1.  Solve  xy  +  if  =  133 (1) 

*a  -  f  =     95 (2) 

Divide  (2)  by  (1)  and  cancel  the  common  factor  x  +  y, 

and  substituting  in  (1) 

W  +  2/2  =  133- 

Solving,  we  get  ?/  =   ±  7  ; 

and  by  substituting  in  (3),      x  =   ±12. 

Note. — This  includes  the  case  where  one  equation  is  exactly 
divisible  by  the  other. 

2.  Solve  cc8  +  yz  =  ISxy (1) 

x  +  y  =  12 (2) 

Divide  (1)  by  (2),    x2  -  xy  +  y2  =  %xy. 

...     x*  _  |.,y  +  y2  =  0 (3) 

Subtract  (3)  from  the  square  of  (2), 
%xy  =  144. 

.-.     \xy=  16 (4) 

Add  (3)  and  (4)  and  take  the  square  root, 

x  —  y  =   ±4    .     .     .     .     »     .     .   (5) 
From  (2)  and  (5)  x  =  8,  or  4, 

?/  =  4,  or  8. 

3.  Solve  x*  +  ®y  +  2/4  =  2613 (1) 

a2  +  xy  +  y2  =  07  ....     .    (2) 

Divide  (1)  by  (2),   x2  -  xy  +  ?/2  =      39 (3) 

Add  (2)  and  (3),               x2  +  y2  =  53. 

Subtract  (3)  from  (2) ,              a^  =  14. 

.-.     05   =  ±    7,     or     ±2, 

2/   =  ±   2,     or     ±7. 


31G  SIMULTANEOUS    QUADRATIC  EQUATIONS. 

Solve  the  following : 

4.    x2  +  xy  =  84,  ^4?is. 


a*2  — 

2/2  =  24. 

5. 

x*  + 

XY  +  y  = 

133, 

x*  + 

a^  +  y9  = 

19. 

6. 

a;3  + 

f  =  407, 

a;   + 

y  =    li. 

7. 

£C     — 

y  =     4, 

a;3  - 

?/3  =  988. 

a;  = 

±7, 

y  = 

±5. 

X  = 

±3, 

or  ±   2, 

y  = 

±2, 

or  ±   3. 

a;  = 

7, 

or        4, 

2/  = 

4, 

or        7. 

x  = 

11, 

or  —   7, 

y  = 

7, 

or  -11. 

149.  Case  IV.  When  the  Two  Equations  are  of 
the  Second  Degree  and  Homogeneous. 

First  method. 

1.    Solve  2a;2  +     3xy  +       v/2  =  70    .     .     .     .   (1) 

Gar  +      xy  —       if  —  50    .     .     .     .   (2) 

Divide  (1)  by  (2),^  +     3**  +       ^      h 
6x2  +       xy  -       y2 

Hence  10a;2  +  loxy  -f     by"  =  42a;2  +  Ixy  —  ly- ; 

or  32a:2  -     8a#  -  12/  =  0. 

This  is  a  quadratic  equation  which  we  may  solve,  and  find 
the  value  of  one  unknown  quantity  in  terms  of  the  other. 
Thus,  solving  for  x, 

x2  -  \xy  =  |?/2.        

By  Art.  13G,  Rule  II,  x  «  |  ±  0//2  +  g  =  Mi/2, 
or  x  =  y    H    y  =  fy,  or  — $y. 

Take  a;  =  |y,  and  substitute  in  either  (1)  or  (2),  and  we 
have  ?/2  =  1G. 

.-.     y  =  ±4,     and     x  =  ±3. 

If  we  substitute  the  second  value  of  a*,  which  is  —  ■?,//,  we 
find  an  inadmissible  result. 

Second  method.  Examples  of  this  class  are  conveniently 
solved  by  substituting  for  one  uf  the  unknown  quantities  the 


SIMULTANEOUS    QUADRATIC  EQUATIONS.  317 

product  of  the  other  and  a  third  unknown  quantity,  called 
an  auxiliary  quantity. 

2.  Solve  2y2  -  Axy  +  3x2  =  17  .  '  .     .     .     .   (1) 

f  -    *2  =  16 (2) 

Put  y  =  vx,  and  substitute  in  (1)  and  (2).     Thus 

x2{2v2  -  4v  +  3)  =  17 (3) 

x2(  v2  -  1)  =  16 (4) 

_     ,.  .  .                      2^2  -4^  +  3         17 
By  division,  2  _   =  ti- 

.-.     32v2  -  64v  +  48    =  17v2-17; 
or  15u2  -  Mv  =  -65. 

Solving,  we  obtain  v  =  f ,  or  -1/. 

Take  v  =  |,  and  substitute  in  either  (3)  or  (4). 
From  (4)       o;2(2¥5-  -  1)  =  16;         .-.     x2  =  9. 

.-.     x  =  ±3,     and     y  =  vx  =  %x  =  ±5. 
Again,  take  v  =  \3-,  x2(  W   -  1)  =  16;         .'.     z2  =  ^. 
.-.     x  =  ±f,     and     y  —  vx  =  ±±£. 
Any  pair  of  equations  which  are  o/  £/te  second  degree  and 
homogeneous,   can  be  solved   by  either  of   these  methods, 
though  the  second  is  usually  preferred. 
Solve  the  following : 

3.  x2  +  Sxy  =  28,  Ans.  x  =  ±4,  or  T 14, 
xy    +  Ay2  =     8.  y  =  ±1,  or  ±   4. 

4.  x2  +    a#  +  2?/2  =  74,  x  =  ±8,  or  ±   3, 
2z2  +  2xy  +    ?/2  =73.                       2/  =  q:5,  or  ±   5. 

5.  a;'2  +    a$  -    6y2  =  24,  SB-  ±6, 
a?  +  3^  _  1(y  _  32.                                      y  =  ±2. 

6.  a;2  +    xy  -  6/  =21,  x  =  ±9, 

a^  —  2?/2  =4.  2/  =  ±4. 

Note.  —  Many  examples  of  homogeneous  equations  of  the  second 
degree  are  easily  solved  by  Case  II  or  III.  Only  those  examples  of 
this  class  are  to  be  solved  by  Case  IV  that  rannot  be  solved  by  either 
Case  II  or  III. 


318  SYMMETRICAL  EQUATIONS. 

150.  Case  V.  When  the  Two  Equations  are 
Symmetrical  with  respect  to  x  and  /.  —  An  expres- 
sion is  said  to  be  symmetrical  zvith  respect  to  two  letters  ivhen 
these  letters  are  similarly  involved,  i.e.,  when  they  can  be 
interchanged  loithout  altering  the  expression.  Thus,  the 
expression  a3  +  a2x  +  ax2  +  x3  is  symmetrical  with  respect 
to  a  and  x,  since  if  we  write  x  for  a,  and  a  for  x,  we  get  the 
same  expression.  Also  x*  +  3x2y  +  3xy2  +  y*  is  symmet- 
rical with  respect  to  x  and  y. 

Examples  of  this  class  may  frequently  be  solved  by 
substituting  for  the  unknown  quantities,  the  sum  and  differ- 
ence of  two  others. 

1.  Solve  x*  +  ?/4  =  82 (1) 

x   -y   =    2 (2) 

Put  x  =  u  +  v,     and     y  =  u  —  v  ; 

(2)  becomes        (u  +  v)  —  {u  —  v)  =     2  ;         .-.     v  =  1. 

(1)  becomes     (u  +  l)4  -f  (u  -  l)4  =  82. 

.-.     2(?t4  +  Git2  +  1)  =  82; 
or  u*  +  Gu2  -  40  =     0. 

Hence,  Art.  139,  (a2  +  10)  (w8  -  4)  =     0. 

.-.     u2  =       4,  or  -10. 

.-.     u    =  ±2,  or  ±\J^To. 
Thus  x    =  3,  -1,       1  ±  \/-10, 

2/  =  i,  -3,  -1  ±  y/^To. 

2.  Solve  (a2  +  r)  (.r3  +  /)  =  280     .     .     .     .   (1) 

a;  +  y  =      4     ....   (2) 
Put  x  =  w  +  v,     and     y  =  w  —  v  J 

(2)  becomes       (tt  +  v)  +  (w  —  v)  =  4  ;         .*.     u  =  2. 
Also     /-  +  //-  =  (2  +  v)2  +  (2  -  v)2  =       8  +     2va, 

and  a3  +  yA  =  (2  +  v)a  +  (2  -  u)3  =     16  +  12v2. 


SPECIAL   METHODS.  319 

Hence  (1)  becomes  (8  +  2v2)(16  +  12v2)  =  280, 

or  (4  -(-  v2)  (4  +  3v2)  =    35. 

...      v*  +  3$.tf  =     Ig-. 

.-.     v2=  -f  ±  li  =  ^  or-y. 
.-.     v  =  ±1,  or  ±y/--139-- 
.-.     a;   =      3,  1,  2  ±  ^-¥> 
?/   =      1,  3,  2  T  \J^K°. 
Solve  the  following : 

0.  x  —  y  —  2,     and     x5  —  ?/5  =  242. 

ylws.  a;  =  3,  or  — 1 ;     ?/  =  1,  or  —3. 

4.  x  —  y  =  1,     and     a5  —  y5  =  781. 

.4ns.  a;  =  4,  —3  ;     y  =  3,  —4. 

5.  x  +  y  =  3,     and     x5  +  ?/5  =  33. 

.<4.7is.  x  =  1,  2;     ?/  =  2,  1. 

151.  Special  Methods.  —  The  preceding  cases  will  be 
sufficient  as  a  general  explanation  of  the  methods  to  be 
employed  ;  but  in  some  cases  special  artifices  are  necessary. 
One  that  is  often  used  with  advantage  consists  in  consider- 
ing the  sum,  difference,  product,  or  quotient,  of  the  two 
unknown  quantities  as  a  single  quantity,  and  first  finding  its 
value.  Other  artifices  may  also  be  used  with  advantage,  but 
familiarity  with  them  can  be  obtained  only  by  experience. 

1.  Solve   x2  +  4xy  +  3a:  =  40  -  6y  -  4y2    .     .     .   (1) 

2xy  -  x2  =  3 (2) 

From  (1)  we  have  x2  +  4xy  +  4y2  -{-  ox  -\-  Gy  =  40  ; 
or  (x  +  2y)2  +  3 (a;  +  2y)  -  40  =  0. 

Consider  x  -f  2y  as  a  single  unknown  quantity,  and  find 
its  value  from  this  quadratic.     Thus, 
(Art.  139),  [(as  +  2y)  +  8] [(as  +  2y)  -  5]  =  0. 

.-.     x  +  2y  =  -8, (3) 

or  x  +  2y  =       5 (4) 


320  SPECIAL   METHODS. 

From  (2)  and  (4)  we  obtain 

x  =  1,  orf  ; 

y  =  2,  or  J. 
From  (2)  and  (3)  we  obtain 

_  -4  ±  y/To 
2  ' 

v  =  -UTt/lO 
*  2 

2.  Solve  ay  -  6fB  =  34  -  3# n) 

3xy  +  y  =  18  +  2x (2) 

Multiply  (2)  by  3  and  subtract  the  result  from  (1), 
xhf  -  dxy  +  20  =  0. 
.-.     (xy  -  5)(xy  -4)  =  0. 

.'.     xy  =  5    .     .     .     .   (3) 
xy  =  4    .     .     .     .   (4) 
From  (2)  and  (3)  we  obtain 

x  =  1,  or  -f, 
y  =  5,  or  -2. 
From  (2)  and  (4)  we  obtain 

x  =  ~3  ±  V/l7,      and      2/  =  3  ±  ^V7. 

Solve  the  following : 

3.  x1  +  y  =  73  -  3a  -  2xy,     Ans.  x  =  4,  10,  - 12  ±  ^58, 
y1  +  a  =  44  -  3y.  *      ?/=5,  -7,  -1  ^  y/58. 

4.  ^  +  ^  =  12,  x=  0,    ^, 

y~      y 

x  -  y  =  3.  ?/  =  3,-?. 

5.  a2  +  3.t?/  =  54,  x  =  ±3,  ±36, 
xy  +  4y'2  =115.                                       y  =   ±5,  T'-V1- 

G.    x4  -  x1  +  .V4  -  ?/2  =  S4,  .r  =   ±3,  ±2, 

.c-  +  x-y'1  +  y1  =  49.  y  =   ±2,  ±3. 


PROBLEMS  LEADING    TO    QUADRATIC  EQUATIONS.   321 

152.  Simultaneous  Quadratic  Equations  with 
Three  Unknown  Quantities. 

1.  Solve  xy  +  xz  =  27 (1) 

yz  +  yx  =  32 (2) 

zx  +  zy  =  35 (3) 

Add  (1)  and  (2)  and  subtract  (3)  from  the  sum, 

2xy  =  24;         .-.     xy  =  12  .     .     .     .   (4) 

Subtract  (4)  from  (1),      xz  =     15 (5) 

Subtract  (4)  from  (2),     yz  =    20 (6) 

Multiply  (4)  and  (5),    x'hjz  =180 (7) 

Divide  (7)  by  (G),  x2  =       9       .-..«  =  ±3. 

Heuce  from  (4),  y  =     12  h-  (±3)  =   ±4. 

And  from  (5),  z  =     15  -f-  (±3)  =  ±5. 

Thus  x  =  ±3,  ?/  =  ±4,  2  =  ±5, 

all  the  upper  signs  being  taken  together. 
Solve  the  following : 

2.  3yz  +  22KB  —  4o^  =16,  -4ws.  x  =  ±1, 
2y«  —  3z.c  +  xy  =  5,  ?/  =  ±2, 
4*/z  —    Z£  —  3.t?/  =  15.                                    2  =    ±3. 

3.  G(x-  +  2/2  +  «2)  =  13(0!  +  y  +  z)  =  ±£l,  a?  =  f ,  f, 
a;?/  =  z2.  2/  =  f ,  |, 

z  =    ±2. 

153.  Problems  Leading  to  Simultaneous  Quad- 
ratic Equations. 

1.  The  small  wheel  of  a  bicycle  makes  135  revolutions 
more  than  the  large  wheel  in  a  distance  of  2G0  yards  ;  if  the 
circumference  of  each  were  one  foot  more,  the  small  wheel 
would  make  27  revolutions  more  than  the  large  wheel  in  a 
distance  of  70  yards  :  find  the  circumference  of  each  wheel. 

Let  x  =  the  circumference  of  the  small  wheel  in  feet, 
and      y  =  the  circumference  of  the  large  wheel  in  feet. 

Then  the  two  wheels  make  —  and  - —  revolutions  respec- 
x  y 

tively  in  a  distance  of  2G0  yards. 


322   PROBLEMS   LEADING    TO    QUADRATIC   EQUATIONS. 


Hence 

780  _ 

X 

780  _ 

y 

135; 

X 

y 

A   • 

Similai 

•iy 

from 

the  second  condition, 

210 

210 

=  27; 

x  +  i     y  +  i 

i  i 


(i) 


(2) 


x  +  1       y  +  1 

52w  ,    -,       61w  +  52 

From  (1),  x-  =  "'      ;       •••     x  +  1  =  „•    ,    -     • 

9y  +  52  9y  +  o2 

Substituting  in  (2),  ^__  _  __  =  ^. 

.-.     9/  -USy  =  52. 
Solving,  we  obtain        y  =  13,  or  — |. 

Substituting  y  =  13  in  (1),  we  find  a;  =  4.  The  negative 
value  of  y  is  inadmissible. 

Hence  the  small  wheel  is  4  feet,  and  the  large  wheel  is  13 
feet  in  circumference. 

2.  A  man  starts  from  the  foot  of  a  mountain  to  walk  to 
its  summit ;  his  rate  of  walking  during  the  second  half  of  the 
distance  is  half  a  mile  per  hour  less  than  his  rate  dining 
the  first  half,  and  he  reaches  the  summit  in  5£  hours.  He 
descends  in  3|  hours  by  walking  at  a  uniform  rate,  which  is 
one  mile  per  hour  more  than  his  rate  during  the  first  half  of 
the  ascent :  find  the  distance  to  the  summit,  and  the  rates 
of  walking. 

Let  2fl3  =  the  number  of  miles  to  the  summit, 
and        y  =  the  rate  of  walking,  in  miles  per  hour,  during 
the  first  half  of  the  ascent. 

Then    -  =  the  time  in  hours  for  the  first  half  of  the  ascent ; 

y 

an(j  — flJ —  =  the  time  in  hours  for  the  second  half  of  the 

y  -  \ 

ascent. 


PROBLEMS   LEADING    TO    QUADRATIC   EQUATIONS.    323 

Hence  -  +  -^-r  =  H 0) 

Similarly  -^  =  3j (2) 

From  (2)  x  =  -^ (>/  +  1) (3) 

From  (1)       x(2y  -  \)  =  h}y{y  -  \) (4) 

Substituting  (3)  in  (4), 

V-0/  +  i)(2y-i)=  V*(y-i). 

.-.     28/  -  89#  =  -15. 
Solving,  we  obtain  y  =  3,  or  2\. 

Substituting  y  =  3  in  (3),  we  find  a;  =  ^5.  The  other 
value  of  y  is  inapplicable,  because  by  supposition  y  is  greater 
than  ^. 

Hence  the  whole  distance  to  the  summit  is  15  miles,  and 
the  rates  of  walking  are  3,  2|,  and  4  miles  per  hour. 

3.  The  sum  of  the  squares  of  two  numbers  is  170,  and 
the  difference  of  their  squares  is  72  :  find  the  numbers. 

Ans.  11  ;  7. 

4.  The  product  of  two  numbers  is  108,  and  their  sum  is 
twice  their  difference  :  find  the  numbers.  Ans.  6  ;  18. 

5.  The  product  of  two  numbers  is  6  times  their  sum,  and 
the  sum  of  their  squares  is  325  :  find  the  numbers. 

Ans.  10  ;  15. 

6.  A  certain  rectangle  contains  300  square  feet ;  a  second 
rectangle  is  8  feet  shorter,  and  10  feet  broader,  and  also 
contains  300  square  feet :  find  the  length  and  breadth  of  the 
first  rectangle.  Ans.  20  ;  15. 

7.  Find  two  numbers  such  that  their  sum  may  be  39,  and 
the  sum  of  their  cubes  17199.  Ans.  15  and  24. 

8.  The  product  of  two  numbers  is  750,  and  the  quotient 
of  one  divided  by  the  other  is  3*  :  find  the  numbers. 

Ans.  50  and  15. 


324  EXAMPLES. 

EXAMPLES    OF    SIMULTANEOUS    QUADRATICS. 

Note.  —  In  the  great  variety  of  simultaneous  quadratic  equations, 
it  is  impossible  to  give  rules  for  every  solution.  The  artifices  employed 
in  Algebraic  work  are  very  numerous.  The  student  is  cautioned  not 
to  go  to  work  upon  a  pair  of  equations  at  random,  but  to  study  them 
until  he  sees  how  they  can  be  reduced  to  a  simpler  equation  by 
addition,  multiplication,  factoring,  or  by  some  other  process,  and  then 
to  perform  the  operations  thus  suggested. 

Solve  the  following : 

1.  x  +  Ay  =  14, 
f  +  Ax  =  2y  +  11. 

2.  fix  +  2y  =  16, 

xy  =  10. 

3.  x    +  2y    =  9, 
3y2  —  5x2  =  43. 

4.  3x  —    y  =  11, 
3a;2  -    y'2  =  47. 

5.  Ax  +  9y  =  12, 
2a;2  +  xy  =     6y\ 

6.  Sx  +  2y  =  5xy, 
15a;  —  4?y  =  4xy. 

7.  a;  +     ?/  =     51, 

xy  =  518. 

8.  x  —    y  =       18, 

a;?/  =  1075. 

9.  a;2  +    y2  =  89, 

xy  =40. 

10.  a;2+     ?/2=  178, 
x  +    2/  =     1G. 

11.  a;2+    ?/2  =     185, 
x  -    y  =        3. 


L»s.  a?  = 

2, 

-4G, 

Z/  = 

3, 

15. 

x  = 

2, 

V. 

y  = 

5, 

3. 

X  = 

1, 

-ft. 

y  = 

4, 

w- 

X  = 

4, 

7, 

y  = 

1, 

10. 

a;  = 

-24, 

f 

y  = 

12, 

t- 

a;  = 

h 

o, 

2/  = 

h 

0. 

a;  = 

37, 

14, 

?/  = 

14, 

37. 

a;  = 

43, 

-25, 

y  = 

25, 

-43. 

x  = 

±    8, 

±   5, 

y  = 

±    5, 

±   8. 

X  = 

18, 

3, 

y  = 

3, 

13. 

X  = 

U, 

-  8, 

y  = 

-s, 

-11. 

EXAMPLES.  325 

12.  x1  —  xy  +  y*  =  7G,  ^Ins.  a;  =  10,        4, 
x  +  y  =  14.  y  =  4,      10. 

13.  x  -  y  =     3,  » =  7,  -   4, 
a2  —  3^2/  +  y1  =  -19.  y  =  4,-7. 


14. 

4  +  -2  =  ttt, 

or       ?/- 

a  = 

f> 

3 
8' 

-  +  -  =  u- 

x        y 

2/  = 

8' 

#• 

15. 

-ji  H r,  =  /(iff' 

a;?/   =  30. 

a;  = 

±   G, 

±   5, 

2/  = 

±   5, 

±   6. 

16. 

x1  +  y2  +  xy  =  208, 

OJ  — 

12, 

4, 

a  +  y  =     16. 

y  = 

4, 

12. 

17. 

^      _      yl     _                  K]^ 

a;  = 

5, 

a  —  y  =        2. 

2/  = 

3. 

18. 

a;3  —  y3  =        7xy, 

a,*  = 

4, 

-   2, 

x   —  y  =        2. 

2/  = 

2, 

-  4. 

19. 

a;   +  y  =       23, 

a;  = 

14, 

9, 

a;3  +  y3  =  3473. 

y  = 

9, 

14. 

20. 

aj    +  y    =       35, 

x  = 

8, 

27, 

aj^  +  y$  =         5. 

2/  = 

27, 

8. 

21. 

33    -  y   =  \/x  +  V^i 

a;  = 

16, 

9, 

a;i  —  y'  =  37. 

2/  = 

9, 

16. 

22. 

a;4  _j_  ^2  _j_  ^4  _  2923, 

a;  = 

±    7, 

±   3, 

a;2  —  xy     -f  ?/2  =       37. 

2/  = 

±    3, 

±    7. 

23. 

a;4  +  kV  +  2/4  =  9211, 

as  = 

±   9, 

±   5, 

a;2  —  xy     +  ?/2  =       61. 

2/  = 

±   5, 

±   9- 

24. 

Xs  -  y3  =  56, 

x  = 

4, 

-   2, 

ar  +  xy  +  2/2  =  28. 

y  = 

2, 

-  4. 

25. 

x3  +  y3  =  126, 

X  = 

5, 

1, 

ar  -  xy  +  y*  =  21. 

2/  = 

1, 

5. 

326  EXAMPLES. 

26.  —  +  —  =  1t^tt,  Ans.  x  =        5,        1, 
x3       y3 

-  +  -  =  if  y  =      i,      s. 

a;       y 

27.  a;2  +  a:?/  =  45,  y2  +  a-v/  =  36.  x=  ±   5  ;  y=  ±4. 

28.  2ar  -xy  =  56,  2xy  —  y2  =  48.  .<•=  ±    7  ;  y=  ±  6. 

29.  x2  —  2xy  =  15,  a*?/  —  2y-  =  7.  ;/■=  ±15  ;  y=  ±  7. 

30.  x*y(x+y) =80,  x*y(2x— 3y)=80.       a-=  ±   4;y=±l. 

31.  a;3  +  1  =  %,  a;  =  2,  $,  -1, 
ar1  -f  a;  =  %y.  y  =  1,  a,      0. 

Note. —  It  will  be  seen  that  Examples  17  to  31  can  be  solved  by 
Case  III. 

32.  ar  +  3xy  =  54,  Ans.  x  =  ±3,  ±36, 
xy  +  Ay2  =115.  y  =   ±5,  T%3-. 

33.  a,-2  +  a;i/  =  24,  x  =  ±4,  ±6V^2, 
2/  +  3xy  =  32.  2/  =  ±2,  =F8V^2. 

34.  a;2  -  3xy  =  10,  x  =  ±5,  ±4, 
4?/2  -  xy  =  -1.  ?/  =  ±1,  ±£. 

35.  a;2  +  xy  -  2y2  =  -44,  x  =  ±14,  ±1, 

xy  +  3?/2  =       80.  y  =  ^    8,  ±5. 

36.  a;2  +  3a;?/  =     54,  x  =  ±3,  ±36, 
xy  +  4y2  =  115.  y  =  ±5,  Tllf 

37.  a?(a;  ±  ?y)  =  40,  x  =  ±5,  ±4V/2, 

y(*  -  y)  =  6.  y  =  ±3,  ±  V2. 

38.  a;(a;  +  ?/)  +  ?/(.r  -  ?/)  =  158,  a-  =  ± !),  ±8^2, 

7x(x  +  y)=    72y(x-y).      ?/=±7,  ±  V2. 

39.  a-  +  y  =  4,  a.-4  +  y4  =      *•_>.  a;  =  3,  1 ;  y  =  1,  3. 

40.  x  —  y  =  3,  a;5  —  ?/'  =  3093.       x  =  5,  —  2  ;  y  =  2,  —5. 

41.  afy3  +  13.tv/  -f  12  =  0,  x  =      4,-3,  L=_V5. 


aj  +  y  =  1. 


y=_S,      4,LT^. 


EXAMPLES.  327 

42.  x  +  y  =    5,  Ans.  x  =      6,  —  1,  5  ±  *17, 

4a$  =  12  -  a;2?/2,  y  =  -1,      G,  5  T  *17. 

43.  xY  +  5a;?/  =  84,  a-  =  7,  1,  4  ±  \^8, 

a;  +  y  =    8.  y  =  1,  7,  4  T  V§8. 

44.  ar  +  4?/2  +  80  =  15a  +  30?/,  a;  =  4,  3,  6,  2, 

xy  =     6.  7/  =  |,  2,  1,  3. 

45.  9a;2  +  ?/2  -  63a;  -  21?/  +  128  =  0,       x  =  2,  |,  4,    -|, 

an/  =  4.       y  =  2,  6,  1,  12. 

46.  a"  +  2/4  =  14a,-2/,  a;  =  5(1  ±  Vs),  |/l  ±  -±=\ 
-   +f-«-  ^2(1=FV/3)^(1T^) 

a^  -  a;  —  y  =  54.  y  =  12,  -9. 


48. 

x3  +  y(xy  -  1)  =  0, 

aj=  V*(^§  -  1), 

y*  -  x(xy  +  1)  =  0. 

1 

V2(\/2  -  1) 

49. 

x5  +  y°  =  an/(aj  +  ?/)3, 

x  =  fVlO  ±  ^V5, 

a-?/4  =  (a;  +  y)3. 
(x«+l)y  =  (y*+l)x*, 

y  =  fV2  ±  |V5.  . 

50. 

a;=i[V/|3/3  +  3  +  V/r3-l], 

(2/6+l)a;  =  9(ar+l)?/3. 

ylns.  ?/  =  i [if 3  •  V^3  +  3  ± V^s/9-  1], 

51.  a;  +  y  +  «  =  13,  x  =  5,  2,  3  ±  V^l, 
a-"2  +  ?/a  +  z~  =  65,  ?/  =  2,  5,  3  T  V^T, 

xy  =10.  2  =  6,  7. 

52.  (x  +  ?/)(a-  -f-  z)  =  30,  as  =  4,  -4, 
(2/  +  2)0/  +  a,")  =  15,  y  =  1,  -1, 
(2  +a?)(2  4-y)  =  18.                                 z  =  2,  -2. 


328  EXAMPLES. 

53.    if  +  yz  +  z1  =  49,  Ans. 

z'2  +  zx  +  a;'-  =  19, 
x2  +  xy  +  if  =  39. 


X  = 

=  ±2, 

V 

y  = 

=  ±5, 

*» 

2!    = 

=  ±3, 

** 

54.   x{y  +  z)  = 


=         /(g  +  ft-c)(a  +  c-fr) 
V  2(6  +  c  -  a) 


x       7  .,  /(«  +  6  —  c)  (6  +  c  —  a) 

/      ,      x  .  /(&  +  c  —  a)  (a  +  c  —  6) 

55.  The  product  of  two  numbers  is  60  times  their  differ- 
ence, and  the  sum  of  their  squares  is  244  :  find  the  numbers. 

Ans.  10,  12. 

56.  Find  two  numbers  whose  difference  added  to  the 
difference  of  their  squares  is  14,  and  whose  sum  added  to 
the  sum  of  their  squares  is  26.  Ans.  4,  2. 

57.  Find  two  numbers  such  that  twice  the  first  with  three 
times  the  second  may  make  60,  and  twice  the  square  of  the 
first  with  three  times  the  square  of  the  second  may  make 
840.  Ans.  18  and  8,  or  6  and  16. 

58.  Find  two  numbers  whose  sum  is  nine  times  their 
difference,  and  whose  product  diminished  by  the  greater 
number  is  equal  to  twelve  times  the  greater  number  divided 
by  the  less.  Ans.  5,  4. 

59.  Find  two  numbers  whose  difference  multiplied  by  the 
difference  of  their  squares  is  32,  and  whose  sum  multiplied 
by  the  sum  of  their  squares  is  272.  Ans.  5,  3. 

60.  Find  two  numbers  whose  product  is  equal  to  their 
sum,  and  whose  sum  added  to  the  sum  of  their  squares  is  12. 

Ans.  2,  2. 

61.  Find  two  numbers  whose  sum  added  to  their  product 
is  34,  and  the  sum  of  whose  squares  diminished  by  their 
sum  is  42.  Ana.   1,  6. 


EXAMPLES.  329 

G2.  A  number  consisting  of  two  digits  has  one  decimal 
place ;  the  difference  of  the  squares  of  the  digits  is  20,  and 
if  the  digits  be  reversed,  the  sum  of  the  two  numbers  is  11 : 
find  the  number.  Arts.  G. 4,  or  4.6. 

G3.  A  man  has  to  travel  a  certain  distance  ;  and  when  he 
has  traveled  40  miles  he  increases  his  speed  2  miles  per 
hour.  If  he  had  traveled  with  his  increased  speed  during 
the  whole  journey  he  would  have  arrived  40  minutes  earlier ; 
but  if  he  had  continued  at  his  original  speed  he  would  have 
arrived  20  minutes  later.  Find  the  whole  distance  he  had  to 
travel,  and  his  original  speed.  Ans.  GO,  10. 

64.  A  and  B  are  two  towns  situated  18  miles  apart  on  the 
same  bank  of  a  river.  A  man  goes  from  A  to  B  in  4  hours, 
by  rowing  the  first  half  of  the  distance  and  walking  the 
second  half.  In  returning  he  walks  the  first  half  at  the  same 
rate  as  before,  but  the  stream  being  with  him,  he  rows  1£ 
miles  per  hour  more  than  in  going,  and  accomplishes  the 
whole  distance  in  3h  hours.  Find  his  rates  of  walking  and 
rowing.  Ans.  4|  walking,  4|  rowing  at  first. 

65.  A  and  B  run  a  race  round  a  two  mile  course.  In  the 
first  heat  B  reaches  the  winning  post  2  minutes  before  A. 
In  the  second  heat  A  increases  his  speed  2  miles  per  hour, 
and  B  diminishes  his  as  much  ;  and  A  then  arrives  at  the 
winning  post  2  minutes  before  B.  Find  at  what  rate  each 
man  ran  in  the  first  heat.  Ans.  10,  12  miles  per  hour. 

66.  Find  two  numbers  whose  product  is  equal  to  the  dif- 
ference of  their  squares,  and  the  sum  of  their  squares  equal 
to  the  difference. of  their  cubes.  Ans.  £05,  -*-(5  +  \/5). 

67.  The  fore- wheel  of  a  coach  makes  6  revolutions  more 
than  the  hind-wheel  in  going  120  yards  ;  but  if  the  circum- 
ference of  each  wheel  be  increased  1  yard,  the  fore-wheel 
will  make  only  4  revolutions  more  than  the  hind-wheel  in  the 
same  distance  :  find  the  circumference  of  each  wheel. 

Ans  4  and  5  yards. 

68.  A  vessel  is  to  be  filled  with  water  by  two  pipes.  The 
first  pipe  is  kept  open  during  three-fifths  of  the  time  which 


330  EXAMPLES. 

the  second  would  take  to  fill  the  vessel ;  then  the  first  pipe 
is  closed  and  the  second  is  opened.  If  the  two  pipes  had 
both  been  kept  open  together  the  vessel  would  have  been 
filled  6  hours  sooner,  and  the  first  pipe  would  have  brought 
in  two-thirds  of  the  quantity  of  water  which  the  second  pipe 
really  brought  in.  How  long  would  each  pipe  alone  take  to 
fill  the  vessel?  Ans.  15  and  10  hours. 

69.  The  number  of  men  in  both  fronts  of  two  columns  of 
troops,  A  and  B,  where  each  consisted  of  as  many  ranks  as 
it  had  men  in  front,  was  84  ;  but  when  the  columns  changed 
ground,  and  A  was  drawn  up  with  the  front  that  B  had,  and 
B  with  the  front  that  A  had,  then  the  number  of  ranks  in 
both  columns  was  91 :  find  the  number  of  men  in  each 
column.  Ans.  2304,  1296. 

70.  Two  trains  start  at  the  same  time  from  two  towns, 
and  each  proceeds  at  a  uniform  rate  towards  the  other  town. 
When  they  meet  it  is  found  that  one  train  has  run  108  miles 
more  than  the  other,  and  that  if  they  continue  to  run  at  the 
same  rate  they  will  finish  the  journey  in  9  and  16  hours 
respectively.  Find  the  distance  between  the  towns  and  the 
rates  of  the  trains.  Ans.  756,  36,  27. 

71.  Two  travelers,  A  and  B,  set  out  from  two  places,  P 
and  Q,  at  the  same  time  ;  A  starts  from  P  with  the  design 
to  pass  through  Q,  and  B  starts  -from  Q  and  travels  in  the 
same  direction  as  A.  When  A  overtook  B  it  was  found  that 
they  had  together  traveled  30  miles,  that  A  had  passu  1 
through  Q  4  hours  before,  and  that  B,  at  his  rate  of  travel- 
ing, was  9  hours'  journey  distant  from  P.  Find  the  dis- 
tance between  P  and  Q.  Ans.  6  miles. 

72.  Two  travelers,  A  and  B,  set  out  at  the  same  time 
from  two  places,  P  and  (,),  respectively,  and  travel  so  as  to 
meet.  When  they  meet  it  is  found  that  A  has  traveled  SO 
miles  more  than  15,  and  that  A  will  reach  Q  in  4  days,  and 
I>  will  reach  P  in  9  days,  after  they  meet,  Find  the  distance 
between  P  and  Q.  Ana.  150  miles. 


RATIO  —  DEFINITIONS.  331 


CHAPTER    XVI. 

RATIO  —  PROPOR  T  ION  —  VARIATION. 

154.  Ratio  —  Definitions. — The  relative  magnitude  of 
two  quantities,  measured  by  the  number  of  times  which  the 
first  contains  the  second,  is  called  their  Ratio. 

The  ratio  of  a  to  b  is  usually  written  a :  b ;  a  is  called 
the  first  term,  and  b  the  second  term  of  the  ratio.  The  first 
term  is  often  called  the  antecedent,  and  the  second  term  the 
consequent. 

Magnitudes  must  always  be  expressed  by  means  of  num- 
bers; and  the  number  of  times  which  one  number  contains 
the  other  is  found  by  dividing  the  one  by  the  other.     Hence 

the  ratio  a  :  b  may  be  measured  by  the  fraction  -. 

Thus,  the  ratio  a :  b  is  equal  to  -,  or  is  -. 

b  b 

Concrete  quantities  of  different  kinds  can  have  no  ratio  to 
one  another ;  thus,  we  cannot  compare  pounds  with  yards, 
or  dollars  with  days. 

To  compare  two  quantities,  they  must  be  expressed  in 
terms  of  the  same  unit.  For  example,  the  ratio  of  4  yards 
to  15  inches  is  measured  by  the  fraction 

4  x  3  x  12  _  48 
15"  ~  "  "*■ 

A  ratio  is  called  a  ratio  of  greater  inequality,  of  less  in- 
equality, or  of  equality,  according  as  the  antecedent  is  greater 
than,  less  than,  or  equal  to  the  consequent. 

Eatios  are  compounded  by  multiplying  together  the  ante- 
cedents of  the  given  ratios  for  a  new  antecedent,  and  the 


332  RATIO  —  DEFINITIONS. 

consequents  for  a  new  consequent.     Thus,  the    ratio   com- 
pounded of  the  three  ratios, 

3a  :  2b,  4ab  :  5c2,  c  :  a, 

is  3a  X  tab  x  c  :  2b  x  5c2  x  a,  or  6a :  5c. 

When  the  ratio  a :  b  is  compounded  with  itself,  the  result- 
ing ratio  is  a2 :  b2,  and  is  called  the  duplicate  ratio  of  a  :  b. 
Similarly,  the  ratio  a3 :  b8  is  called  the  triplicate  ratio  of  a :  b. 
Also  the  ratio  a* :  b*  is  called  the  swbduplicate  ratio  of  a :  b. 

If  we  interchange  the  terms  of  a  ratio,  the  result  is  called 
the  inverse  ratio.     Thus 

b  :  a  is  the  inverse  of  a :  b. 

The  inverse  ratio  is  the  reciprocal  of  the  direct  ratio. 

When  the  ratio  of  two  quantities  can  be  exactly  expressed 
by  the  ratio  of  two  integers,  the  quantities  are  said  to  be 
commensurable;  when  the  ratio  cannot  be  exactly  expressed 
by  the  ratio  of  two  integers,  they  are  said  to  be  incommen- 
surable. 

Although  we  cannot  find  two  integers  which  will  exactly 
measure  the  ratio  of  two  incommensurable  quantities,  yet  we 
can  always  find  two  integers  whose  ratio  differs  from  the 
required  ratio  by  as  small  a  quantity  as  we  please. 

For  example,  the  ratio  of  a  diagonal  to  a  side  of  a  square 
cannot  be  exactly  expressed  b}T  the  ratio  of  two  whole 
numbers,  for  this  ratio  is  \J2,  and  we  cannot  find  any 
fraction  which  is  exactly  equal  to  \J2  ;  but  by  taking  a  suf- 
ficient number  of  decimals,  we  may  find  \J2  to  any  required 
degree  of  approximation.     Thus 

\]2  =  1.4142135 

and  therefore     \J2  >  -\*  ft  •}  J  •>  and  <  \  g  ' ■;',  jj  J  < . 
That  is,  the  ratio  of  a  diagonal  to  a  side  of  a  a  square  lies 
between  f$<VouiVo  ;U1,1    !  ,m <m,' , <', •  ;lll(l  therefore  differs    from 
either  of  these  ratios  by  less  than  one-millionth;  and  since 
the  decimals  may  be  continued  without  end  in  extracting  the 


PROPERTIES   OF  RATIOS.  333 

square  root  of  2,  it  is  evident  that  this  ratio  can  be  expressed 
as  a  fraction  with  an  error  less  than  any  assignable  quantity. 

In  general.  When  a  and  b  are  incommensurable,  divide  b 
into  n  equal  parts  each  equal  to  x,  so  that  b  =  nx,  where  n 
is  a  positive  integer.  Also  let  a  >  mx,  but  <  (m  +  l)x, 
then 

a     mx       ,      (m  4-  l)x 

-  >  —  and  <  ^ ! —  ; 

.  b      nx  nx 

that  is,  -  lies  between  —  and  — — —  ;  so  that  -  differs  from 
b  n  n  h 

—  by  a  quantity  less  than  -.     And  since  we  cau  choose  x 
(our  unit  of  measure)  as  small  as  we  please,  n  can  be  made 

as  great  as  we  please,  and  therefore  -  can  be  made  as  small 

n 

as  we  please.     Hence  two  integers,  m  and  n,  can  be  found 

whose  ratio  will  express  the  ratio  a  :  b  to  any  required  degree 

of  accuracy. 

Note.  —  The  student  should  observe  that  the  Algebraic  definition 
of  ratio  deals  with  numbers,  or  with  magnitudes  represented  by 
numbers,  while  the  Geometric  definition  of  ratio  deals  with  concrete 
magnitudes,  such  as  lines  or  areas  represented  Geometrically,  but  not 
referred  to  any  common  unit  of  measure. 

155.  Properties  of  Ratios. —  (1)  If  the  terms  of  o. 
ratio  be  multiplied  or  divided  by  the  same  number  the  value  of 
the  ratio  is  unaltered. 

-c,  a       ma  , .   ,    „nx 

For  -  =  —  (Art.  79). 

b        mb 

Thus  the  ratios  2:3,  6:9,  and  2m :  3m,  are  all  equal  to 
each  other. 

Two  or  more  ratios  are  compared  by  reducing  the  fractions 
which  measure  them  to  a  common  denominator.  Thus,  sup- 
pose a, :  b  and  c  :  d  are  two  ratios.  Then  -  =  — ,  -  =  —  ; 
1  b        bd    d        bd 

hence  the  ratio  a  :  b  >,  = ,  or  <  the  ratio  c :  d,  according  as 
ad  >,  = ,  or  <  be. 


334  PROPERTIES   OF  RATIOS. 

The  ratio  of  two  fractions  can  be  expressed  as  a  ratio  of 

two  integers. 

a 

Thus  the  ratio  -:-  is  measured  by  the   fraction  -  or  —  ; 
b   d  J  c         be 

and  is  therefore  equivalent  to  the  ratio  ad :  be. 

(2)  A  ratio  of  greater  inequality  is  diminished,  and  a  ratio 
of  less  inequality  is  increased,  by  adding  the' same  quantity  to 
each  <>f  its  terms;  that  is,  the  ratio  is  made  more  nearly  equal 
to  unity. 

Let  a :  b  be  the  ratio,  and  let  a  +  x :  b  +  x  be  the  new 
ratio  formed  by  adding  x  to  each  of  its  terms. 

Then  *  -  ±±*  =  »(°  ~  &>  ; 

b        b  +  x        b{b  +  x) 

and  a  —  b  is  positive  or  negative  according  as  a  is  greater 

or  less  than  b. 

TT  a  +  x  a  ,.  7      ,,    .  . 

Hence <,  or  >  -    according,  as  a  >,  or  <  b  ;  that  is, 

b  +  x  b 

the  resulting  ratio  is  brought  nearer  to  unity. 

For  example,  if  to  each  term  of  the  ratio  3  :  2  we  add  12, 
the  new  ratio  15  :  14  is  less  than  the  former,  because  ||  =  l^ 
is  clearly  less  than  §  =  1^. 

Also,  if  to  each  term  of  the  ratio  2  :  3  we  add  12,  the  new 
ratio  14:1;")  is  greater  than  the  former,  since  \4  is  clearly 
greater  than  f. 

(3)  Similarly,  it  can  be  proved  that  a  ratio  of  greater  in- 
equality is  increased,  and  a  ratio  of  less  inequality  is  dimin- 
ished, by  taking  the  same  quantity  from  both  its  terms. 

(4)  The  following  is  a  very  important  proposition  concern- 
ing equal  ratios. 

If  r  =  -  =  —  = ,  then  each  of  these  ratios 

b       d       f 

_  /pa"  +  qcn  +  ren  +  .  .  .  A" 

\/>t>"  +  qdn  +  rf"  + /  ' 

where  p,  q,  r,  n  are  any  quantities  whatever. 


EXAMPLES.  335 

=  Je ; 


6       d       / 

then     a  =  &ft,  c  =  dk,  e  =  fk ; 

therefore  pa* +q<?+ren+  .  .  .  .  =})buk"  +  qd"k" +rf"kn+ 


/pa"  +  q<?  +  re"  + \» 


fc==^  =  ^  =  !=(l) 
b      d     J 

By  giving  different  values  to  p,  g,  r,  n,  many  particular 
cases  of  this  general  proposition  may  be  deduced  ;  or  they 
may  be  proved  independently  by  the  above  method. 

Suppose  n  =  1,  then  we  have  from  (1) 

a  _  c  _  e  _                 _  pa  +  7C  +  r(J  +  •  •  •  •        t%\ 
b~  d~  f      pb  +  qd  +  rf  + 

Suppose  »=  1,  and  p=q=r= ,  then  (1)  becomes 

a  _  c  _  e  _              _  ft  +  c  +  e  .  .  .  .  ,^\ 

b~d~f~  b  +  d  +  / 

That  is,  ?o/>e?t  «  sencs  of  fractions  are  equal,  each  of  them  is 
equal  to  the  .sum  of  all  the  numerators  divided  by  the  sum  of 
all  the  denominators. 


EXAMPLES. 

1.  If  -  =  i,  find  the  value  of  bf  ~  3Jf. 
y  7x  +  -ly 

OX  q 

hx  —  Sy  _   y  _  *£  —  3 


7a;  +  -ly       7x       g        V"  +  2 
2/ 
2.  If  a :  &  be  in  the  duplicate  ratio  of  a  +  a: :  b  +  as,  prove 
that  :«2  =  o&. 


From  the  condition 


/a  +  a\2  _  « 

a-b  +  2abx  +  &.ea  =  a&2  +  2a&sc  +  ax2. 
,-.     x-  =  ab. 


336  PROPORTION  —  DEFINITIONS. 

Find  the  ratio  compounded  of 

3.  The  ratios  i  :  15  and  25  :  30.  Aus.  5  :  27. 

4.  The  ratio  27  :  8,  and  the  duplicate  ratio  of  4  :  3.      6:1. 

5.  The  ratio  10'J  :  200,  and  the  duplicate  ratio  of  15 :  2G. 

Arts.  9  :  32. 

6.  If  Ax"  +  if  =  4x77,  fiud  the  ratio  &  :  y.  1:2. 

7.  What  is  the  ratio  x :  ?/,  if  the  ratio  Ax  ■+■  by :  3x  —  y  is 
equal  to  2?  Ans.  7:2. 

8.  If  lx  —  Ay :  3x  +  y  =  b  :  \o,  find  the  ratio  x  :  >/. 

Ans.  3:4. 

PROPORTION. 

156.  Definitions. — Four  quantities  are  said  to  be  in 
proportion  when  the  ratio  of  the  first  to  the  second  is  equal 
to  the  ratio  of  the  third  to  the  fourth  ;  and  the  terms  of  the 
ratios  are  said  to  be  projiortionals. 

Thus,  if  -  =  -,  then  a,  b,  c,  d,  are  called  proportionals, 
b       d 

or  are  said  to  be  in  proportion.     The  proportion  is  written 

a  :  b  —  c :  d, 
or  a:  b  ::  c  :  d, 

which  is  read  "  a  is  to  &  as  c  is  to  d." 

The  Algebraic  test  of  a  proportion  is  that  the  two  fractions 
which  represent  the  ratios  shall  be  equal. 

The  four  terms  of  the  two  equal  ratios  are  called  the  terms 
of  the  proportion.  The  first  and  fourth  terms  are  called  the 
extremes,  and  the  second  and  third,  the  means.  Thus,  in 
the  above  proportion,  a  and  d  are  the  extremes  and  b  and  c 
the  means. 

Quantities  are  said  to  be  in  continued  proportion  when  the 
first  is  to  the  second,  as  the  second  is  to  the  third,  as  the 
third  to  the  fourth,  and  so  on.  Thus  a,  b,  c,  d,  e,  /,  .  .  . 
are  in  continued  proportion  when 

a  :  b  =  b  :  c  =  c  :  d  =  d  :  e  =  e  :  /  = 


PROPERTIES   OF  PROPORTIONS.  337 

If  a,  b,  c,  be  in  continued  proportion,  b  is  said  to  be  a 
mean  proportional  between  a  and  c ;  and  c  is  said  to  be 
a  tMrd  proportional  to  a  and  b. 

If  a,  b,  c,  d  be  in  continued  proportion,  b  and  c  are  said 
to  be  two  mean  jJroportionals  between  a  and  d  ;  and  so  on. 

157.  Properties  of  Proportions. —  (1)  If  four  quan- 
tities are  in  proportion,  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means. 

Let  the  proportion  be  a  :  b  =  c:  d. 

Then  by  definition  (Art.  156),  -  =  -. 

Multiplying  by  bd,  ad  =  be (1) 

Hence  if  any  three  terms  of  a  proportion  are  given,  the 
fourth  may  be  found  from  the  relation  ad  =  be. 

Note.  —  This  proposition  furnishes  a  more  convenient  test  of  a 
proportion  than  the  one  in  Art.  156.  Thus,  to  ascertain  whether 
2  :  5  : :  6  :  16,  it  is  only  necessary  to  compare  the  product  of  the  means 
and  extremes;  and  since  5  x  6  is  not  equal  to  2  x  16,  we  see  that 
2,  5,  6,  16,  are  not  in  proportion. 

If  b  =  c,  we  have  from  (1),  ad  —  62;     .*.  b  =  \ac. 

That  is,  the  mean  proportional  between  tioo  given  quanti* 
ties  is  equal  to  the  square  root  of  their  product. 

(2)  Conversely,  If  the  product  of  two  quantities  be  equal 
to  the  product  of  tioo  others,  two  of  them  may  be  made  the 
extremes,  and  the  other  two  the  means,  of  a  proportion. 

For  let  ad  =  be. 

Dividing  by  bd,  -  =  - ; 

b        d 

that  is,  a  :  b  :  :  c  :  d. 

In  a  similar  manner  it  may  be  shown  that  the  proportions 


arc: 

:   b:d, 

b :  a  : 

:  d:  c, 

b:d  : 

:   a:  c, 

c:  d  : 

:  a:b,  etc.. 

are  all  true  provided  that  ad  = 

be. 

338  PROPERTIES   OF  PROPORTIONS. 

If  four  quantities  are  in  proportion  they  icill  be  in  propor- 
tion by 

(3)  Inversion.  —  If  a  :  b  :  :  c  :  d,  then  h  :  a  :  :  d  :  c. 

For  -  =  - ;  therefore  1  +  ?  =  1  -*-  C-  \ 

b       d  b  d 

that  is,  -  =  -  ;  or  & :  a  : :  d :  c. 

a       c 

(4)  Alternation.  —  If  a :  b  :  :  c:d,  then  a  :  c  ::  bid. 

For  ad  =  be;  therefore  —  =  — - ; 
cd       cd 

that  is  -  =  -  ;  or  a  :  c  :  :  b  :  cl. 
c        d 

(5)  Composition.  —  If  a :  & : :  c:  d,  then  a+6 : 6 : :  c+d:  d. 

For  -  =  - ;  therefore  j  +  1  =  C-  +  1 ; 
6       d  b  d 

that  is  ^±J>  =  2J±J  ;ora  +  &:&::c  +  d:d. 

&  d 

(G)  Division.  —  If  a :  &  : :  c :  d,  then  a  —  & :  & : :  c  —  d:d. 

For  -  =  -  ;  therefore  ~  -  1  =  3  -  1 ; 
&       d  b  d 

that  is  a  ~  6  =  ^^  ;  or  a  -  6 :  6'  >•:  c  -  d :  d. 

6  d 

In  a  similar  manner  it  may  be  shown  that  the  sum  (or  the 
difference)  of  the  first  and  second  of  two  quantities  is  to  the 
first  as  the  sum  (or  the  difference)  of  the  third  and  fourth 
is  to  the  third. 

(7)  Composition  and  Division.  — If  a:b  : :  c:  d,  then 

a  +  6 :  a  —  b  : :  c  +  d :  e  —  d. 
For  by  (5)  and  (6), 

a  +  b       c  +  rZ        a  —  &  _  c  — <? . 


b               d                  b               d 

by  division, 

a  +  b  __c  +  d. 

0  —  b       c  —  d 

or 

a  +  b:  a  —  b  :  :  C  +  d  i  G  —  d. 

PROPERTIES   OF  PROPORTIONS.  339 

(8)  If  three  quantities  are  in  continued  proportion,  the  first 
is  to  the  third  in  the  duplicate  ratio  of  the  first  to  the  second. 

For  if  a  :  b  :  :  b  :  c,  then  -  =  -. 
b        c 

■^       a       a    ,  b       a  „  a       a2 

Now  -  =  -  X  -  =  7  X  -  =  - . 

c        b        c       b        b       b- 

Hence  a  :  c  : :  a2  :  b2. 

Similarly  it  may  be  shown    that    if    a :  b  ::  b  :  c  : :  c :  d, 

then  aid  :  :  a3 :  b3. 

(9)  Quantities  which  are  proportional  to  the  same  quanti- 
ties, are  proportional  to  each  other. 

If  a :  6  : :  e :  /,  and  c:  d: :  e:  f,  then  a  :  b  :  :  c  :  d. 

-r-,      a        e         i  c        e      .,        ~       a        c 
For  -  =  — ,  and  -  =  - ;  therefore  -  =  -, 

b       f  d       /'  6       d'        . 

or  a:b::  c  :  d. 

(10)  T/ie  products  of  the  corresponding  terms  of  tioo  or 
more  proportions  are  in  proportion. 

For  if  a  :  b : :  c :  d,  and  e:  f  ::  g:h, 

a       c  ■,      e       a 

then  -  =  -,     and     -  =  f 

6       d  /A 

therefore  —  =  -^  ;     or     ae  :bf::cg:  dh. 

of       dh 

(11)  When  four  quantities  are  in  proportion,  if  the  first 
und  second  be  multiplied,  or  divided,  by  any  quantity,  as 
also  the  third  and  fourth,  the  resulting  quantities  will  be  in 
■proportion. 

For  if  a :  b : :  c :  d,  then  -  =  -  ; 
b       d 

therefore        —  =  —  ;     or     ma  :  mb  :  :  nc  :  nd. 
mb       nd 

Similarly  it  may  be  shown  that  —  :  —  ::-:-. 
m     m       n     n 


340         PROPERTIES   OF  PROPORTIONS. 

(12)  In  a  similar  manner  it  may  be  shown  that  if  the  first 
anil  third  terms  be  multiplied,  or  divided,  by  any  quantity, 
and  also  the  second  and  fourth,  the  resulting  quantities  will 
be  in  proportion. 

(13)  If  four  quantities  are  in  x>i'oportion,  the  like  powers, 
or  roots,  of  these  quantities  will  be  in  proportion. 

For  if  a :  6 : :  c :  d,  then  -  =  -  ; 
b       d 

a"        c" 
therefore         —  =  —  :         .-.     a"  :  bn : :  c"  :  dn. 
6"       d" 

fin  r~n  X         1  X  L 

Also  —i  =  —;         .-.     a*:bn::cn:d*. 

b«      d» 

(14)  If  any  number  of  quantities  are  in  proportion,  any 
antecedent  is  to  its  consequent,  as  the  sum  of  all  the  antecedents 
is  to  the  sum  of  all  the  consequents. 

For  if  a  :  b  :  :  c  :  d  :  :  e  :  f, 
then  by  (1),  ad  =  6c,  and  af  =  be  ;  also  ab  =  ba. 
Adding  a(b  +  d  ■+-  /)  =  b(a  +  c  +  e); 

therefore  by  (2),a:6::a  +  c  +  e:6  +  d  +  /. 

This  also  follows  directly  from  (3)  of  Art.  155. 

(15)  When  -,-,—, are  unequal,  it  follows  from 

b   d   f 

Ex.  3  of  Art.  10G,  that 

a  +  c  +  e  +  g  + 


b+d+f+h+ ■ 

is  greater  than  the  least,  and  less  than  the  greatest,  of  the 

..       ..        a    c     e    q 

fractions  -,-,—,, 

b   d   f   h 

It  is  obvious  from  the  preceding  propositions  that  if  four 

quantities  are  in  proportion,  many  other  proportions  may  be 

derived  from  them.     The  propositions  just   proved  are  often 

useful   in  solving   problems.      In  particular,  the  solution  of 

certain  equations  is  greatly  facilitated  by  a  skilful  use  of  the 

operations  of  composition  and  division. 


EXAMPLES.  341 

EXAMPLES. 
1.    If  a  :  b  ::  c  :  d, 

show  that  a2  +  ab  :  c2  +  cd  : :  62  —  2ab  :  da  —  2cd. 

Let  -  =  -  =  x ;  then  a  =  6#,  and  c  =  d.t\ 
b       d 

a2  +  a&  _.  62t- 2  +  62o;  =  & 

•'*     ca  +  Cd        rfV2  +  d2aj       d2' 

62  -  2a6  _  b2  -  2b2x  _  62 
da  -  2cd       d2  -  2d2a;       d2' 
Therefore  by  (9),  a2  +  ab  :  c2  +  cd  : :  62  -  2ab  :  d2  -  2cd. 
If  3a  +  6b  +  c  +  2d  _  3a  +  66  —  c  —  2d 
3a  —  66  +  c  —  2d       3a  —  66  —  c  +  2d' 
prove  that  a  :  &  : :  c  :  d. 

Bv  m  2 (3a  +  c)    =    2 (3a  -  c) 

*  k  ;  2(66  +  2d)        2(66  -  2d)' 

r  r  /4x  3a  +  c  _  66  +  2d 

y  ^  '  3a  -  c  ~  66  -  2d' 

Again  by  (7),  6a  =  126  ^     0  .  6  . .  c  .  rf. 

&         J  v   7  2c        4d 

3.  Find  a  fourth  proportional  to  x3,  xy,  bx'hj.        Ans.  by2. 

4.  Find  a  mean  proportional  between  \2cix2  and  3a3. 

Ans.  Ga2x. 
•      5.    Find  a  third  proportional  to  x3  and  ?«2.  4.x. 

6.    If  a  :  6  :  :  c  :  d,  show  that 

(1)  ac  :  6d  ::  c2:  d2. 

(2)  a6  :  cd   ::  a2:  c2. 

•        (3)   a2   :  c2    :  :  a2  -  b2  :  c?  -  d*. 
7.    If  a  :  6  :  :  c  :  d,  prove  that 

( 1 )  a6  +  cd :  «6  -  cd : :  a'2  +  c2  f  a2  -  c2. 

(2)  a2  +  ac  +  c'2  :  a2  -  ac  +  c2  : :  62  +  &d  +  d2 :  6'2  -  6d  +  d2 

(3)  a  :  6  :  :  s/Sa2  +  5c2  :  v'362  +  5d2. 

(4)  a  +  6 :  c  +  d : :  Va2  +  62 :  V^2  +  d2. 


342  VARIATION  —  DEFINITION. 

8.    If  at  :  b  : :  c  :  d  : :  e  :  /,  prove  that 

2a9  +  .V  -  56s  :   2&2  +  3d2  -  5/2  : :  ae  :  &/. 

.r2  +  x  —  2        \.r  +  bx  — 


9.    Solve  the  equation 


-  2  5a;  -  6 

Ans.  x  =  0,  —2. 

10.  Find  a;  in  terms  of  y  from  the  proportions  x:  y  : :  a3 :  b3, 
and  a  :  b  : :  \Jc  +  x  :  Vd  +  y.  ^       fC  =  £ 

11.  If  a,  fr,  c,  d  are  in  continued  proportion,  prove  that 

a  :  d  : :  a3  +  &3  +  c3  :  &3  +  c3  +  d3. 

VARIATION. 

158.  Definition.  —  One  quantity  is  said  to  vary  directly 
as  another  when  the  two  quantities  depend  upon  each  other 
in  such  a  manner  that  if  one  be  changed  the  other  is  changed 
in  the  .same  proportion. 

Thus,  if  a  train  moving  uniformly,  travels  40  miles  in  an 
hour,  it  will  travel  80  miles  in  2  hours,  120  miles  in  :i  hours. 
and  so  on  ;  the  distance  in  each  case  being  increased  or 
diminished  in  the  same  ratio  as  the  time.  This  is  expressed 
by  saying  that  when  the  velocity  is  uniform,  the  distance  is 
proportional  to  the  time,  or  more  briefly,  the  distance  varies 
as  the  time.  We  may  express  this  result  with  Algebraic 
symbols  thus:  let  A  and  a  be  the  numbers  which  represent 
the  distances  traveled  by  the  train  in  the  times  represented 
by  the  numbers  B  and  b  ;  that  is,  when  A  is  changed  to  any 
other  value  a,  B  must  be  changed  to  another  value  6,  so  that 
A  :  a  '.'.  B  :  &;  then  A  is  said  to  vary  directly  as  B,  or 
more  briefly,  to  vary  as  B. 

Another  phrase,*  which  is  also  in  use,  is  '■'■A  is  proportional 
to  B." 

*  Strictly  Bpeaklog,  this  phrase  is  better  than  the  one  "varies  as,"  which  is 
somewhat  antiquated ;  but  in  deference  to  usage  we  retain  it.  The  student  must  not 
suppose  that  the  variation  hrVr  considered  is  the  only  kind.  We  are  not  here 
concerned  with  variation  in  gent  rat,  but  merely  « i 1 1 ■.  the  simplest  of  all  the  possible 
kimi»  of  variation. 


DIFFERENT   CASES   OF    VARIATION.  343 

This  relation  is  sometimes  expressed  by  the  symbol  oc,  so 
that  A  oc  B  is  read  "A  varies  as  B." 

It  will  thus  be  seen  that  variation  is  merely  an  abridged 
method  of  expressing  proportion,  and  that  four  quantities 
are  understood  though  only  two  are  expressed. 

If  A  varies  as  B,  then  A  is  equal  to  B  multiplied  by  some 
constant  quantity. 

For  suppose  that  or,  av  «2,   .  .  .  .  ,  b,  bv  &2, are 

corresponding  values  of  A  and  B. 

Let  a  and  b  denote  one  pair  of  these  values,  so  that  when 
A  has  the  value  a,  B  has  the  value  b  ;  then  we  have  by  the 
definition,  A  :  a  : :  B  :  b.     Hence 

A  =  -B  =  mB, 

b 

where  m  is  equal  to  the  constant  ratio  a  :  b. 

159.  Different  Cases  of  Variation.  —  There  are  four 
different  kinds  of  variation. 

(1)  One  quantity  is  said  to  vary  Directly  as  another  when 
the  two  increase  or  decrease  together  in  the  same  ratio. 
Thus. 

A  oc  B,  or  A  =  mB  (Art.  158). 

For  example,  If  a  man  works  for  a  certain  sum  per  hour, 
the  amount  of  his  wages  varies  as  the  number  of  hours 
during  which  he  works. 

(2)  One  quantity  is  said  to  vary  Inversely  as  another  when 
the  first  varies  as  the  reciprocal  of  the  other.  Thus  A  varies 
inversely  as  B  is  written 

A  oc  — ,  or  A  ■=  — ,  where  m  is  a  constant. 
B  B 

For  example,  If  a  man  has  to  perform  a  certain  journey, 
the  time  in  which  he  will  perform  it  varies  inversely  as  his 
speed.  If  he  doubles  his  speed,  he  will  go  in  half  the  time  ; 
and  so  on. 


344  PROPOSITIONS   IN    VARIATION. 

(3)  One  quantity  is  said  to  vary  as  two  others  Joint};/, 
when  the  first  varies  as  the  product  of  the  other  two.  Thus 
A  varies  as  B  and  C  jointly  is  written 

A  oc  BC,  or  A  =  mBC,  where  m  is  a  constant. 

For  example,  The  wages  to  be  received  by  a  workman  will 
vary  as  the  number  of  days  he  has  worked  and  the  tvecges  per 
clay  jointly. 

(4)  One  quantity  is  said  to  vary  Directly  as  a  second  and 
Inversely  as  a  third,  when  it  varies  jointly  as  the  second  and 
the  reciprocal  of  the  third.  Thus  ^1  varies  directly  as  B 
and  inversely  as  C  is  written 

A      B         ,  B      , 

A  oc  — ,  or  A  =  m  — ,  where  m  is  a  constant. 
C  C 

For  example,  The  base  of  a  triangle  varies  directly  as  the 
area  and  inversely  as  the  altitude. 

In  the  different  cases  of  variation  just  defined,  to  deter- 
mine the  constant  m  it  will  only  be  necessary  to  have  given 
one  set  of  corresponding  values. 

Example  1.    If  Ace  B,  and  A  =  3  when  B  =  12,  we  have 

A  =  mB ;         .-.     3    =  m  x   12; 

or  m  =  £  ;  .  • .     A  =  \  B. 

2.  If  A  varies  as  B  and  inversely  as  0,  and  A  —  6  when 
B  =  2  and  C  =  9,  we  have 


A  =  in    r ; 

.-.     6   =  m  x 

in  =  27  ; 

A            0~B 

,.     .1  =  2,-, 

160.  Propositions  in  Variation.  —  The  simplest  method 
of  treating  variations  is  to  convert  them  into  equations. 

(1)  If  A  cc  B,  and  5  cc  C,  then  A  cc  C. 

For  let       A  =  mil,  and  B  =  wC  (Art.  158), 
where  m  and  ft  are  constants. 


PROPOSITIONS   IN    VARIATION.  345 

Then  A  =  mnC ; 

.•.     A  <x  C,  since  win  is  constant. 

In  like  manner,  if  A  cc  B,  and  B  oz  -,  then  vl  oc  — . 

G  0 

(2)  If  A  cc  0,  and  B  cc  C,  then  .1  ±  B  oc  C,  and  \l  AB  cc  C. 
For  let  A  —  mC,  and  23  =  nO, 

where  m  and  n  are  constants. 

Then  ^L  ±  B  =  (m  ±  n)C. 

.-.     A  ±  B  cc  C,  since  m  ±  «  is  constant. 
Also  \lA~B  =  \JmnC1  =  CVwwi. 

.-.     ^123  cc  C,  since  v«»i  is  constant. 

(3)  If  A  cc  BC,  then  23  cc  4  and  C  oc  — . 

C  B 

For  let  ^  =  mBC;  then  23  =  -  ^. 

...     23  cc^.       Similarly  C  oc  |. 

(4)  If  J.  cc  B,  and  0  cc  Z>,  then  AC  cc  Si). 
For  let  A  =  m23,  and  C  =  w2). 

Then  AC  =  mnBD.         .-.     AC  oz  BD. 

(5)  If  ^1  cc  J3,  then  u4"  cc  23". 

For  let  .4  =  m23 ;  then  An  =  mnB". 

.-.     yincc23\ 

(6)  If  A  oz  B  when  C  is  constant,  and  A  cc  C  when  B  is 
constant,  then  A  <x  BC  when  both  B  and  C  are  variable. 

The  variation  of  A  depends  on  the  variations  of  the  two 
quantities  B  and  C.  Suppose  these  latter  variations  to  take 
place  successively,  each  in  its  turn  producing  its  own  effect 
on  A. 

Let  then  B  be  changed  to  b,  and  in  consequence  let  A  be 
changed  to  a',  C  being  constant ;  then,  by  supposition, 
A  =  B 
a'        b ' 


346  PROPOSITIONS   IN    VARIATION. 

Now  let  C  be  changed  to  c,  and  in  consequence  let  a?  be 
changed  to  a,  b  being  constant;  then,  by  supposition, 
a'       C 


Aa:BC. 


a 

c 

a        a 

-fxf 

b         c 

Hence 

a'        a 

a         be 

The  following  are  illustrations  of  this  proposition. 

The  amount  of  work  done  by  a  given  number  of  men  varies 
directly  as  the  number  of  days  they  work,  and  the  amount 
of  work  clone  in  a  given  time  varies  directly  as  the  number  of 
men;  therefore  when  the  number  of  days  and  the  number 
of  men  are  both  variable,  the  amount  of  work  will  vary  as 
the  product  of  the  number  of  men  and  the  number  of  days. 

Again,  the  area  of  a  triangle  varies  directly  as  the  base 
when  the  height  is  constant,  and  directly  as  the  height  when 
the  base  is  constant;  hence  when  both  the  base  and  the 
height  are  variable,  the  area  will  vary  as  the  product  of 
the  base  and  height. 

In  the  same  manner,  if  A  varies  as  each  of  any  number  of 
quantities,  B,  C,  D,  .  .  .  when  the  rest  are  constant,  then 
when  they  all  vary  A  varies  as  their  product.  Also,  the 
variations  may  be  either  direct  or  inverse. 

Note.  —  This  principle  is  interesting  because  of  its  frequent 
occurrence  in  Physical  Science.  For  example,  in  the  theory  of  gases 
it  is  found  by  experiment  that  the  pressure  p  of  a  gas  varies  as  the 
"absolute  temperature"  t  when  the  volume  v  is- constant,  and  that 
the  pressure  varies  inversely  as  the  volume  when  the  temperature  is 
constant;  that  is, 

p  cc  f,  when  v  is  constant; 

and  p  cc  -,  when  t  is  constant. 

v 

From  these  results,  we  should  expect  that,  when  both  /  and   B  vary. 

We  should  have  the  formula 

P  cc  -,  or*—  =  a  constant, 

v         t 

which  by  actual  experiment  is  found  to  be  the  case. 


PROPOSITIONS   IN    VARIATION.  347 

1.  If  y  varies  inversely  as  or  —  1,  and  is  equal  to  24  when 
x  =  10,  find  y  when  x  =  5. 

Since  7/oc  — - — ,  y  =       m     ,  by  (2)  of  Art.  159. 

J      x2  -  1     J        x1  -  1      J  V   ' 

As  y  —  24  when  x  =  10,  Ave  have 

24  =  — .         .-.     m  =  24  X  99. 
99 

Hence,  when  x  =  5,  we  have 

=  24  x   99  =  99 

a;2  —  1 

2.  The  pressure  of  a  gas  varies  jointly  as  its  density  and 
its  absolute  temperature  ;  also  when  the  density  is  1  and  the 
temperature  300,  the  pressure  is  15.  Find  the  pressure 
when  the  density  is  3  and  the  temperature  is  320. 

Let  p  =  the  pressure,  t  =  the  temperature,  and  d  =  the 
density. 

Then,  since  p  <x  tel,  we  have  p  =  mtd,  by  (3)  of  Art.  159. 
As  p  =  15  when  t  =  300  and  d  =  1,  we  have 
15  =  m  x  300  x  1.         .-.     m  =  fc 
Hence,  when  d  =  3  and  t  =  320,  we  have 
p  =  ^  X  320  X  3  =  48. 

3.  The  time  of  a  railway  journey  varies  directly  as  the 
distance  and  inversely  as  the  velocity  ;  the  velocity  varies 
directly  as  the  square  root  of  the  quantity  of  coal  used  per 
mile,  and  inversely  as  the  number  of  cars  in  the  train.  In  a 
journey  of  25  miles  in  half  an  hour  with  18  cars,  10  cwt.  of 
coal  is  required :  how  much  coal  will  be  consumed  in  a 
journey  of  21  miles  in  28  minutes  with  1G  cars? 

Let  t  =  the  time  in  hours,  d  =  the  distance  in  miles, 
v  =  the  velocity  in  miles  per  hour,  q  =  the  quantity  of  coal 
in  cwt.,  and  n  =  the  number  of  cars.. 

Then  we  have  t  oc  -,  and  v  oc  ^-2. 

v  n 

dn  .  dn 

.'.     t  oc  -— ,     or     t  =  m— . 


Vg  ^q 


348 


EXAMPLES. 


As  d  =  25  when  t  =  £,  n  =  18,  and  ?  =  10,  we  have 
i  =  J**J*.       .     m  =     s/io        and  f  =     y/Io     ^ 
V10  25  x  36  25  x  36y^ 

Hence,  when  d  =  21,  *  =  ff,  and  n  =  1G,  we  have 

28  =  V^JQ  X  21  x  1G  __  y'lQ  x  28 
25  x  36^  25  x  3sfq' 

.-.     V^  =  Vf10x28x60=wio 

25  x  3  X  28  5 

■•-     V  =  V  =  6f. 

Hence  the  quantity  of  coal  is  Gf  cwt. 

4.  A  varies  as  £,  and  A  is  5  when  Z?  is  3  ;   what  is  A 
when  B  is  5?  j„,s#  gx 

5.  J.  varies  inversely  as  B,  and  vl  is  4  when  B  is  15  ; 
what  is  A  when  2?  is  12?  Ans.  5. 

6.  If  x  cc  y  and  y  cc  z,  show  that  az  oc  ?/2. 

7.  If  #  cc  -,  and  y  cc  -,  prove  that  2  cc  x. 

y  z 

8.  If  x  cc  z  and  ?/  cc  z,  prove  that  .<r  —  y"  cc  »2. 

EXAMPLES. 

Find  the  ratio  compounded  of 

1.  The  ratio  32  :  27,  and  the  triplicate  ratio  of  3:4. 

Ans.  1:2. 

2.  The  ratio  6  :  25,  and  the  subdnplicate  ratio  of  25  :  36. 

Ans.  1 :  5. 

3.  The  triplicate  ratio  of  x :  y,  and  the  ratio  2//'2:  3.r\ 

Ans.  2x :  3y. 

4.  If  x-.y  =  8£,  find  the  value  of  (a  -  3y):(2aj  -  by). 

Ans.  1:5. 

5.  If  5  =  J,  and  ^  =  #,  find  the  value  of    3aa?  ~  fo  . 

«  2/  4fy/  —  7aas 

J//S.  17:7. 

6.  Find  x  :  v/,  having  given  a9  -f  6ya  =  bxy.  2  or  3. 


EXAMPLES.  349 

7.  Find  two  numbers  in  the  ratio  of  5  to  6,  and  whose 
sum  is  121.  Ans.  55  and  66. 

8.  For  what  value  of  as  will  the  ratio  15  +  x  :  17  +  x 
be  £?  Ans.  -13. 

9.  Find  £C  in  order  that  x  +  1  :  x  +  4  may  be  the  duplicate 
ratio  of  3:5.  -4ws.  11:16. 

10.  Two  numbers  are  in  the  ratio  of  4  :  5,  and  if  6  be 
taken  from  each,  the  ratio  is  that  of  3  :  4  :  find  the  numbers. 

Ans.  24,  30. 

11.  Find  two  numbers  in  the  ratio  of  5  :  6,  such  that  their 
sum  has  to  the  difference  of  their  squares  the  ratio  of  1  :  7. 

Ans.  35  :  42. 

12.  Find  x  so  that  x  :  1   may  be  the  duplicate  of  the 
ratio  8  :  x.  Ans.  4. 

13.  If  2x :  3y  be  in  the  duplicate  ratio  of  2x  —  m  :  3y  —  ra, 
prove  that  m2  =  6xy. 

14.  If  A :  B  be  the  subduplicate  ratio  of  A  —  x  :  B  —  x, 
prove  that  x  =  AB  :  (A  +  B). 

lo.    Prove  that  if  -1— - — *-  =  -2— ! c  =  -*— J L  ,  each 

of   these   ratios   is   equal   to   1   +  x   :    1   +  y,    supposing 
ax  +  a2  +  a8  not  to  be  zero. 

ir     Tf    a  —  6    _    6  —  c    __    c  —  a    _      a  +  6  +  c 
ay  +  bx       bz  +  ex       cy  +  az        aa  +  6y  +  cz 
prove  that  each  of  these  ratios  =1  :  x  +  y  -f-  2,  supposing 
a  +  6  +  c  not  to  be  zero. 

17.  Find  a  mean  proportional  between  a36  and  «63. 

ylns.  a2b2. 

18.  Find  a  third  proportional  to  (a  —  b)2  and  a2  —  ft2. 

4«s.  (a  +  &)2. 

19.  If  a  :  b  :  :  c  :  d,  prove  that 


(1)  2a  +  3c  :3a  +  2c: 

(2)  At  +  m6  :  pa  +  a/; : 


(3)  Sja1  +  b2:\]c2  +  d2 

(4)  a2c  +  ac'2 :  bhl  +  6d2 


26  +  3d :  36  +  2d. 
Ic  +  md :  pc  +  ad. 


V«3  +  63  :  Vc8  +  d3. 
(a  +  c)3:(6  +  d)3. 


\x 

1 

+  5: 
:  v*  + 

3se-5. 

x  +  4. 
:4a>-l: 

:2, 

2 
5 

or  0, 
or  0 

IX 

-1): 

H 

(2) 

(3) 


350  EXAMPLES. 

Find  the  value  of  a;  in  each  of  the  proportions : 

20.  3x  -  1  :  Gx  -  7 : :  7x  —  10 :  9a  +  10.      Ans.  8  or 

21.  x2-2x  +  B:2x-3::x2- 

22.  2./'  -  1  :.r  +  4::ar  +  2a; 

23.  (V/^TT  +  V/.^I):(V/a7+l 

24.  If  a  :  6 : :  c :  d : :  e :/,  prove  that 

a3  +  c3  +  e3 :  &3  +  d3  +  /3 : :  ace :  bdf. 

25.  If  a  :  b  :  :  c  :  d,  prove  that 

(1)  «(c  +  d)  =  c(a  +  6). 

(«  +  c)(«2  +  c-)  =  (h  +  rQ(ft2  +  d2) 
(a  -  c)(a2  -  c2)       (6  -  d)(&a  -  d2)' 
pa2  +  </»&  +  i'b2   _  pc2  +  qcd  +  rd2 
Za2  +  mab  +  ?<£r        fc2  +  mod  +  nd2" 

26.  If  a  and  ?/  be  unequal  and  x  have  to  ?/  the  duplicate 
ratio  of  x  +  z  :  y  +  z,  prove  that  2  is  a  mean  proportional 
between  a;  and  y. 

27.  If  a  :  &  :  :p  :  g,  prove  that  a2+&2: : :  p2-\-q2:  -± — . 

a+b  p+q 

28.  If  four  quantities  are  in  proportion,  and  the  second  is 
a  mean  proportional  between  the  third  and  fourth,  prove 
that  the  third  will  be  a  mean  proportional  between  the  first 
and  second. 

on     Tj.a  +  b  +  c  +  d       a  +  6  —  c  —  d  ,,    . 

29.  If  — '  ^    7  =  7 -,  prove  that  a, 

a  —  o  -\-  c  —  d      a  —  o  —  c  4-  d 

b,  c,  d  are  in  proportion. 

30.  Each  of  two  vessels  contains  :i  mixture  of  wine  and 
water;  a  mixture  consisting  of  equal  measures  from  the  two 
vessels  contains  as  much  wine  as  water,  and  another  mixture 
consisting  of  four  measures  from  the  first  vessel  and  one 
from  the  second  is  composed  of  wine  and  water  in  the  ratio 
of  2  :  3.  Find  the  proportion  of  wine  and  water  in  each  <>f 
the  vessels. 

Ans.  In  the  first  the  wine  is  J  of  the  whole;  in  the 
second   5. 


EXAMPLES.  351 

31.  If  the  increase  in  the  number  of  male  and  female 
criminals  be  1.8  per  cent.,  while  the  decrease  in  the  number 
of  males  alone  is  4.G  per  cent.,  and  the  increase  in  the 
number  of  females  is  9.8  per  cent.,  compare  the  number  of 
male  and  female  criminals  respectively. 

Ans.  Female  criminals  four-fifths  of  the  male. 

32.  If  xozy,  and  y  =  7  when  a;  =  18,  find  x  when  y  =  21. 

Ans.  54. 

33.  If  x  oc  -,  and  y  =  4  when  x  =  15,  find  y  when  x  =  6. 

y 

Ans.  10. 

34.  A  varies  jointly  as  5  and  G;  and  A  =  G  when  5  =  3 
and  (7=2:  find  A  when  B  =  5  and  C  =  7.  Ans.  35. 

35.  A  varies  jointly  as  B  and  C ;  and  A  =  9  when  5  =  5 
and  C  =  7  :  find  I?  when  ^1  =  54  and  0  =  10.        J.MS.  21. 

36.  A  varies  directly  as  B  and  inversely  as  C ;  and  A  =  10 
when  B  =  15  and  C  =  6 :  find  .1  when  £  =  8  and  C=2. 

Ans.  16. 

37.  If  3a  +  7fr  <x  3«  +  135,  and  a  =  5  when  6  =  3,  find 
the  equation  between  a  and  fr.  Ans.  3«  =  56. 

38.  A  oc  5,  and  yl  =  2  when  5=1;  find  .4  when  B  =  2. 

,4ns.  4. 

39.  If  ^l'2  +  52  oc  A2  —  52,  prove  that  A  +  B  ac  A  —  B. 

40.  3  J.  +  523  cc  54.  +  35 ;  and  A  =  5  when  5  =  2:  find 
the  ratio  A  :  B.  Ans.  5:2. 

41.  A  oc  nB  +  C;  and  .4  =  4  when  5  =  1  and  C  =  2  ; 
and  J.  =  7  when  5  =  2  and  G  =  3  :  find  ».  J.ns.  2. 

42.  If  z  oc  pa;  +  y ;  and  if  z  =  3  when  as  =  1  and  y  =  2, 
and  z  =  5  when  &  =  2  and  ?/  =  3  ;  find  p.  Ans.  1. 

43.  If  x  oc  y  when  z  is  constant,  and  x  oc  -  when  y  is  con- 

z 

stant,  prove  that,  if  y  and  z  both  vary,  a;  will  vary  as  ". 

44.  If  3,  2,  1  be  simultaneous  values  of  x,  y,  z  in  the 
last  example,  determine  the  value  of  x  when  y  =  2  and 
z  =  4.  ^Ins.  f. 


352  EXAMPLES. 

45.  If  x-  oc  y3,  and  x  =  2  when  >/  =  3,  find  the  equation 
between  x  and  y.  Ans.  27a2  =  \y3. 

46.  If  y  varies  as  the  sum  of  two  quantities,  one  of  which 
varies  as  x  directly,  the  other  as  x  inversely,  and  if  y  =  4 
when  x  =  1,  and   y  =  5    when    x  =  2,    find   the    equation 

between  x  and  y.  Ans.  y  =  2x  H — 

x 

47.  If  one  quantity  vary  directly  as  another,  and  the 
former  be  f  when  the  latter  is  f ,  find  what  the  latter  will  be 
when  the  former  is  9.  Ans.  16. 

48.  If  one  quantity  vary  as  the  sum  of  two  others  when 
their  difference  is  constant,  and  also  vary  as  their  difference 
when  their  sum  is  constant,  show  that  when  these  two  quan- 
tities vary  independently,  the  first  quantity  will  vary  as  the 
difference  of  their  squares. 

49.  If  y  =  the  sum  of  two  quantities,  one  of  which  varies 
directly  as  x,  and  the  other  inversely  as  ar ;  and  if  y  =  19 

when  x  =  2,  or  3  ;  find  y  in  terms  of  x.     Ans.  y  =  5a;  -\ — -. 

X' 

50.  If  y  varies  as  the  sum  of  three  quantities  of  which 
the  first  is  constant,  the  second  varies  as  aJ,  and  the  third  as 
or2;  and  if  y  =  0  when  x  =  1,  y  =  1  when  x  =  2,  and 
y  =  4  when  x  =  3  ;  find  y  when  x  =  7.  Ans.  36. 

51.  If  y  =  the  sum  of  three  quantities,  of  which  the  first 
is  constant,  the  second  varies  as  x,  and  the  third  varies  as  a?  ; 
also  when  x  =  3,  5,  7,  y  =  0,  —12,  —32,  respectively;  find 
y  in  terms  of  x.  Ans.  y  —  3  +  2x  —  x'1. 

52.  If  y  varies  as  the  sum  of  three  quantities,  of  which 
the  first  is  constant,  the  second  varies  as  x,  and  the  third  as 
x2 ;  also  when  x  =  a,  2a,  3a,  y  =  0,  a,  4a,  respectively  ; 
show  that  when  x  =  na,  y  =  (n  —  l)2a. 


DEFINITIONS  —  FORM  UL^E.  353 


CHAPTER    XVII. 

ARITHMETIC,     GEOMETRIC,     AND      HARMONIC 
PROGRESSIONS. 

ARITHMETIC    PROGRESSION. 

161.  Definitions  —  Formulae.  —  A  number  of  terms 
formed  according  to  some  law  is  called  a  series.  Quantities 
are  said  to  be  in  Arithmetic  Progression*  when  they  increase 
or  decrease  by  a  constant  difference,  called  the  common 
difference. 

Thus,  the  following  series  are  each  in  Arithmetic  Progres- 
sion : 

2,  5,  8,  11,  14,  17, 

9,  7,  5,  3,  1,  -1,  —3,  —5, 

a,  a  4-  d,  a  +  2(Z,  a  +  3d,  a  +  4(Z, 

The  letters  A.  P.  are  often  used  for  shortness  instead  of 
the  term  Arithmetic  Progression. 

The  common  difference  is  found  by  subtracting  any  term 
of  the  series  from  that  which  immediately  follows  it.  In  the 
first  series  above  the  common  difference  is  3  ;  in  the  second 
it  is  —  2  ;  in  the  third  it  is  d. 

The  series  is  said  to  be  increasing  or  decreasing,  according 
as  the  common  difference  is  positive  or  negative.  Thus,  the 
first  series  above  is  increasing,  and  the  second  is  decreasing. 
If  we  examine  the  third  series  above,  we  see  that  the  coef- 
ficient of  d  in  any  term  is  less  by  one  than  the  number  of  the 
term  in  the  series. 

Thus  the  2d   term  is  a  +  d, 

3d   term  is  a  +  2d, 
4th  term  is  a  +  3d, 

*  Called  also  Arithmetic  Series. 


354  DEFINITIONS  —  FORM  ULJE. 

and  so  on.     Hence  if  n  be  the  number  of   terras,  and  if  I 
denote  the  last,  or  nth  term,  we  have 

I  =  a  +  (n  -  l)d (1) 

Let  S  denote  the  sum  of  n  terms  of  this  series ;  then  we 
have 

8  =  a  +  («  +  d)  +  (a  +  2d)  +...+(*-  2d)  +  (l-  d)  +1 : 
and,  by  writing  the  series  in  the  reverse  order,  we  have 
8  =  I  +  (I  -  d)  +  (I  -  2rf)  +  .  .  .  +  (a  +  2d)  +  (a  +  d)  +  a. 
Adding  together  these  two  equations,  we  have 

2S  =  (a  +  V)  +  (a  +  I)  +  (a  +  0  + to  n  terms 

=  w(a  +  ?). 

.'.     iSf  =  ^(a  +  /) (2) 

By  (1)  and  (2)  we  have  S  =  ~[2«  +  (n  -  l)d]     .     .    (3) 

We  have  here  three  useful  formulae,  (1),  (2),  (3),  which 
should  be  remembered  ;  in  each  of  these  any  one  of  the 
letters  may  denote  the  unknown  quantity  when  the  other 
three  are  known.  For  example,  in  (1),  we  can  write  down 
any  term  of  an  A.  P.  when  the  first  term,  the  common  dif- 
ference, and  the  number  of  the  term  are  given.  Thus,  if  the 
first  term  of  an  A.  P.  is  5  and  the  common  difference  is  3, 
the  10th  term  =  5  +  (10  -  1)3  =  32, 
and  the  20th  term  =  5  +  (20  -  1)3  =  62. 

Also  in  (2),  if  we  substitute  given  values  for  S,  n,  1,  we 
obtain  an  equation  for  finding  a;  and  similarly  in  (3). 
Tims, 

1.  Find  the  sum  of  20  terms  of  the  series  1,  3,  5,  7,  .  .  . 

Here  a  =  1,  d  =  2,  n  =  20;  therefore  by  (3) 

S  =  %°[2  +  11)  X  2]  =  10  x  40  =  400. 

2.  The  first  term  of  a  scries  is  5,  the  last  4;"..  and  the  sum 
400  ;  find  the  number  of  terms,  and  the  common  di  1'1'e  re  nee. 


DEFINITIONS  —  FORM  UL^E.  355 

Here        a  =  5,  I  =  45,  S  =  400 ;  therefore  by  (2) 
400  =  -(5  +  45)  =  25m.         .-.     n  =  16. 

By  (1)  45  =  5  +  15&         .'.     d  =  2f. 

When  any  taco  terms  of  an  A.  P.  are  given,  the  series  can 
be  completely  determined  ;  for  the  data  furnish  two  simul- 
taneous equations,  with  two  unknown  quantities,  which  may 
be  solved  by  -methods  previously  given. 

3.  The  10th  and  15th  terms  of  an  A.  P.  are  25  and  5 
respectively  ;  find  the  series. 

Here  25  =  a  +  9d  ; 

and  5  =  a  +  14tf. 

By  subtraction,  20  =  —  5d.         .*.     d  =  —4. 

Then  a  =  5  —  14cZ  =  61. 

Hence  the  series  is  61,  57,  53, 

4.  Find  the  sum  of  the  first  n  odd  integers. 
Here  a .  =  1,  and  d  =  2  ;  therefore  by  (3) 

S  =  "[2  +  («  -  1)2]  =  ;i  X  2n  =  n2. 
2  2 

Thus  the  sum  of  any  number  of  consecutive  odd  integers 

beginning  with  unity,  is  the  square  of  their  number.* 

Find  the  last  term  and  sum  of  the  following  series : 

5.  14,  64,  414, to  20  terms.  Ans.  964,  9780. 

6.  9,  5,  1, to  100  terms.  -387,  -18900. 

l,  -f , to  21  terms.  -9f,  -99f. 


4'  4' 


Find  the  sum  of  the  following  series  : 

8.  5,  9,  13, to  19  terms.  779. 

9.  12.  9,  6, to  23  terms.  -483. 

Find  the  series  in  which 

10.    The  27th  term  is  186,  and  the  45th  is  312. 

Ans.  4,  11,  18 

*  This  proposition  was  knowu  to  the  Greek  geometers. 


356  ARITHMETIC  MEAN. 

11.  The  9th  term  is 

Am.  1,  -£,  -2, 

12.  The  16th  term  is  214,  and  the  51st  is  739. 

Am.  -11,  4,  19, 

162.  Arithmetic  Mean.  —  When  three  quantities  are  in 
Arithmetic  Progression,  the  middle  one  is  called  the  Arith- 
metic Mean  of  the  other  two. 

Thus  if  a,  b,  c  are  in  A.  P.,  &  is  the  arithmetic  mean  of  a 
and  c  ;  and  by  the  definition  of  A.  P.  we  have 

b  —  a  =  c  —  6 ; 

...     i  =  l(ft  +  C). 

Tims  the  arithmetic  mean  of  any  two  quantities  is  half  their 
sum. 

Between  any  two  given  quantities  any  number  of  terms 
may  be  inserted  so  that  the  whole  series  thus  formed  shall 
be  in  A.  P.  ;  the  terms  thus  inserted  are  called  the  arithmetic 
means. 

For  example,  to  insert  four  arithmetic  means  between  10 
and  25. 

Here  we  have  to  find  an  A.  P.  with  4  terms  between  10  and 
25,  so  that  10  is  the  first  and  25  is  the  sixth  term. 

By  (1)  of  Art.  161, 

25  =  10  +  5cZ;         .-.     d  =  3. 

Thus  the  series  is  10,  13,  16,  19,  22,  25  ; 
and  the  required  arithmetic  means  between  10  and  25  are 
13,  16,  19,  22. 
In  general.     To  insert  n  arithmetic  means  between  a  and  b. 
Here  we  have  to  find  an  A.  P.  with  n  terms  between  a  and 
&,  so  that  a  is  the  first  and  b  is  the  (w  -f-  2)Ul  term. 
By  (1)  of  Art.  161, 

b  =  a  +  (u  +  •>  -  \)d  =  a  -f  (w  +  l)d 

.-.     d  =  b^^. 
n  +  1 


ARITHMETIC    MEAN.  357 

Thus  the  required  means  are 

,    b  —  a  .    -.6  —  a  b  —  a 

a  H ,     a  +  2 , a  +  n 

n  +  1  n  +  1  ?i+l 

1.  Find  the  sum  of  the  first  £>  terms  of  the  series  whose 
wth  term  is  3n  —  1 . 

By  putting  ft  =  1,  and  n  =  p  respectively,  we  obtain 

first  term  =  2,  last  term  =  3p  —  1 . 

Hence  by  (2),  Art.  161,  S  =  2(2  +  3j>  -  1)  =  £(3p  +  l). 

In  an  Arithmetic  Progression  when  a,  S,  and  d  are  given, 
n  is  to  be  found  by  solving  the  quadratic  (3),  Art.  161. 
When  both  roots  are  positive  and  integral,  there  is  no  diffi- 
culty in  interpreting  the  result  corresponding  to  each. 

2.  How  many  terms  of  the  series  24,  20,  16, must 

be  taken  that  the  sum  may  be  72  ? 

Here  a  =  24,  d  =  -4,  S  =  72.  Then  from  (3),  Art. 
161,  we  have 

72  =  5[2  X  24  +  (n  -  l)(-4)]  =  24n  -  2n(n  -  1). 

.-.     n2  -  13n  +  36  =  0,  or  (n  -  4)  (n  -  9)  =  0. 
.-.     ?i  =  4,  or  9. 

Both  of  these  values  satisfy  the  conditions  of  the  ques- 
tion ;  for  if  we  take  the  first  4  terms,  we  get  24,  20,  16,  12  ; 
and  if  we  take  the  first  9  terms,  we  get  24,  20,  16,  12,  8,  4, 
0,  —4,  —  8,  in  either  of  which  the  sum  is  72;  the  last  5 
terms  of  the  last  series  destroy  each  other,  so  that  the  sum 
of  the  first  4  terms  is  the  same  as  the  sum  of  the  first  9 
terms. 

When  one  of  the  roots  is  negative  or  fractional,  it  is  inap- 
plicable, for  a  negative  or  a  fractional  number  of  terms  is, 
strictly  speaking,  without  meaning.  In  some  cases  however 
a  suitable  interpretation  can  be  given  for  a  negative  value 
of  n. 


358  ARITHMETIC  MEAN. 

3.  How  many  terms  of  the  series  —9,-6,-3, 

must  be  taken  that  the  sum  may  he  G6  ? 

Here  66  =  "[-18  +  (n  -  1)8]. 

.-.     ?i2-7w-44  =  0;     or    (n  -  ll)(n  +  4)  =  0. 

.-.     n  =  11,  or  -4. 

If  we  take  11  terms  of  the  series,  we  have 

-9,  -6,  -3,  0,  3,  6,  9,  12,  15,  18,  21  ; 

the  sum  of  which  is  66. 

If  we  begin  at  the  last  term  and  count  backwards  four 
terms,  the  sum  is  also  G6.  From  this  we  see  that,  although 
the  negative  solution  does  not  directly  answer  the  question 
proposed,  we  are  enabled  to  give  it  an  intelligible  meaning 
as  follows  :  begin  at  the  last  term  of  the  series  which  is 
furnished  by  the  positive  value  of  ??,,  and  count  backwards  for 
as  many  terms  as  the  negative  value  indicates  ;  then  the 
result  will  be  the  given  sum.  AVe  thus  see  that  the  negative 
value  for  n  answers  a  question  closely  connected  with  that 
to  which  the  positive  value  applies. 

4.  How  many  terms  of   the  series  26,  21,   16, 

must  be  taken  that  the  sum  may  be  74? 

Here  74  =  -[52  +  (n  -  l)(-5)]. 

Solving,  we  get  n  =  4,  or  7-|. 

Thus,  the  only  applicable  value  of  n  is  4.  Wo  infer  that 
of  the  two  numbers  7  and  8,  one  corresponds  to  a  sum 
greater,  and  the  other  to  a  sum  less  than  71. 

5.  Insert  3  arithmetic  means  between  12  and  -JO. 

A ns.  II.  16,  18. 

6.  Insert  5  arithmetic  means  between  14  and  16. 

-'"*•  II?,.  US 

7.  Insert  17  arithmetic  means  between  98  and  69. 

Arts.  91$,  «»oi, 7,|i. 


GEOMETRIC  PROGRESSION.  359 

How  many  terms  must  be  taken  of  the  series 

8.  42,  30,  3G, to  make  315?      Ans.  14,  or  15. 

9.  _i6, -15,-14, to  make -100?       8,  or  25. 

10.  20,  18f,  17|, to  make  162£?         13,  or  20. 

11.  The  sum  of  three  numbers  in  A.  P.  is  39,  and  their 
product  is  2184  ;  find  them.  Ans.  12,  13,  14. 

12.  The  sum  of  10  terms  of  an  A.  P.,  whose  first  term 
is  2,  is  155  ;  what  is  the  common  difference?  Ans.  3. 

GEOMETRIC    PROGRESSION. 

163.  Definition  —  Formulae.  —  Quantities  are  said  to 
be  in  Geometric  Progression  when  they  increase  or  decrease 
by  a  constant  factor,  called  the  common  ratio. 

Thus,  the  following  series  are  each  in  Geometric  Progres- 
sion (G.  P.): 

3,  6,  12,  24,  48, 

Q      1  lit 

°1    11      "SP       "9'    2ll 

a,  ar,  ar2,  ar3,  aiA, 

The  common  ratio  is  found  by  dividing  any  term  of  the 
series  by  that  which  immediately  precedes  it.  In  the  first 
series  above  the  common  ratio  is  2  ;  in  the  second  it  is  ^ ;  in 
the  third  it  is  r. 

The  series  is  said  to  be  increasing  or  decreasing,  according 
as  the  common  ratio  is  greater  than  1,  or  less  than  1. 
Thus,  the  first  series  above  is  increasing,  and  the  second  is 
decreasing. 

Note  1.  —  An  Arithmetic  Progression  is  formed  by  repeated  addition 
or  subtraction;  a  Geometric  Progression  by  repeated  multiplication  or 
division. 

If  we  examine  the  third  series  above,  we  see  that  the 
exponent  of  r  in  any  term  is  less  by  one  than  the  number  of 
the  term  in  the  series. 

Thus,  the  2d  term  is  ar, 

3d    term  is  ar2, 
4th  term  is  ar3, 


360  GEOMETRIC  PROGRESSION. 

and  so  on.     Hence  if  n  he  the  Dumber  of  terms,  and  if  I 
denote  the  last,  or  utu  term,  we  have 

1=  <"-1 (1) 

Let  S  denote  the  sum  of  n  terms  of  this  series  ;  then  we 
have 

S  =  a  +  ar  +  ar  + +  ar"-3  +  a?-"-1 ; 

multiplying  by  r,  we  have 

SV  =  ar  +  ar2  + +  ar"-2  +  a/-"-1  +  ar*. 

Hence  by  subtraction,  we  have 

Sr  -  S  =  ar*  -  a  ;     or    (r  -  1)S  =  a{r*  -  1). 

s  =  a(f  -  1)    or  «(1  -  r") 

r  -  1  1  -  r  w 

Multiplying- (1)  by  r,  and  substituting  in  (2),  we  get 

c       rl  —  a        a  —  rl  ,ON 

/S  =  ,  or  ,      ....   (3) 

r  -  1  1  -  r  w 

a  form  which  is  sometimes  useful. 

Note  2.  —  It  will  be  found  convenient  to  remember  both  forms 
given  in  (2)  for  S,  and  to  use  the  first  form  in  all  cases  when  r  is 
positive  and  >  1,  and  the  second  when  r  is  negative  or  <  1. 

1.  Find  the  8th  term  of  the  series  —  ■§■,  -|,  —  f , 

Here  a  =  — i,  n  =  8,  r  =  |  h-  (  — ^)  =  —  § :    therefore 

'Jy(l)  /_  lf_3\7__l/_2187\ 

1  —  ~  t\     2)    —      tv. — rsrJ 
=  fit  =  the  8th  term- 

2.  Sum  the  series  1,  3,  9, to  G  terms. 

Here  a  =  1,  n  =  G,  r  =  3;    therefore  by  the  first  form 

°f(2)'  ^  =  3^-j  =  729  -  1  =  3G4j 

3-1  2 

3.  Sum  the  series  81,  54,  36, to  9  terms. 

Here  a  =  81,  v  =  9,  r  =  54  ■—  81  =  §  ;  therefore  by  the 
second  form  of  (2), 

S  =  gll1  ~  (JQ!]  =  248  [1  -  (|)»] 

1  —  r! 


213-  -r  =236|f. 


GEOMETRIC    MEAN.  361 

4.  Sum  the  series  2,  —3,  |,  — to  7  terms. 

Here  a  =  2,  n  =  7,  r  =  —  § ;  therefore  by  the  second  form 

°f  (2)'  S  =  2fl  -  (-I)7]  =  2fl  +  -¥M 

1  -  (-1)  I 

=  3  x  W^  =  utf. 

5.  Find  the  Oth  term  of  each  of  the  following  series : 

(1)   9,  3,  1,  etc.  ;     (2)   2,  -3,  f,  etc.  ;     (3)   a2,  aft,  £/2,  etc. 

Jm.  (l)  ^i    (2)  -W;    (3)|- 

Sum  the  following  series  : 

6.  1,      4,  16, to  6  terms.  Ans.  1365. 

7.  25,     10,    4, to  4  terms.  40f . 

8.  |,  -1,    |, to  7  terms.  -4gV- 

9.  3,-1,    $, to  6  terms.  2|^. 

164.    Geometric  Mean.  —  When  three  quantities  are  in 

Geometric  Progression  the  middle  one  is  called  the  Geometric 
Mean  between  the  other  two. 

Thus  if  a,  b,  c  are  in  G.  P.,  b  is  the  geometric  mean 
between  a  and  c ;  and  by  the  definition  of  G.  P.,  we  have 
6  =  c. 
a       b ' 
b2  =  ac;         .*.     b  =  Sac. 

Thus,  the  geometric  mean  between  any  two  quantities  is  the 
square  root  of  their  product. 

Quantities  which  are  in  G.  P.  are  in  continued  proportion, 
and  the  geometric  mean  between  .two  quantities  is  the  same 
as  their  mean  proportional  (Art.  156). 

Between  any  two  given  quantities  any  number  of  terms  may 
be  inserted  so  that  the  whole  series  thus  formed  shall  be  in 
G.  P.  ;  the  terms  thus  inserted  are  called  the  geometric  means. 

For  example,  to  insert  three  geometric  means  between  2 
and  32. 

Here  we  have  to  find  a  G.  P.  with  3  terms  between  2  and 
32,  so  that  2  is  the  first  and  32  is  the  fifth  term. 

By  (1)  of  Art.  163,  32  =  2r4 ;         .-.     r  =  2. 


3G2     THE   SUM    OF  AN   INFINITE   NUMBER   OF    TERMS. 

Thus  the  series  is  2,  4,  8,  1G,  32,  and  the  required 
geometric  means  between  2  and  32  are  4,  8,  l(i. 

In  general.      To  insert  n  geometric  means  between  a  and  b. 

Here  we  have  to  find  a  G.  P.  with  n  terms  between  a  and 
&,  so  that  a  is  the  first  and  b  is  the  (u  +  2)"1  term. 

By  (1)  of  Art.  1G3, 


b  =  arn+1:         .-.     Vi+1  =  - 


"VI- 


a 

(1) 


Thus  the  required  means  are  ar,  ar2, ar",  where  r 

has  the  value  found  in  (1). 

1.  Insert  4  geometric  means  between  1G0  and  5. 

Ans.  80,  40,  20,  10. 

2.  Insert  6  geometric  means  between  5G  and  — l\. 

Ans.  -28,  14,  -7,  |,  -J,  |. 

3.  Insert  4  geometric  means  between  5£  and  40^. 

Ans.  8,  12,  18,  27. 
165.  The  Sum  of  an  Infinite  Number  of  Terms.  — 
From  (2)  of  Art.  1G3,  we  have 

s  _  «(1  —  r")  _  __a «r^_  _         ^ 

1  -  r  1  -  r       1  -  r 

Now  suppose  r  is  a  proper  fraction,  positive  or  negative  ; 
then  the  greater  the  value  of  n  the  smaller  is  the  absolute 

value  of  r",  and    consequently  of  ;    and  by  taking  n 

sufficiently  large  r"  can   be   made   as   small  as   we  phase. 
Hence,  by  taking  n  large  enough,  the  sum  of  n  terms  of  the 

series  can  be  made  to  differ  from  — —  by  as  small  a  quan- 
tity as  we  please. 

Thus,  the  sum  of  an  infinite  number  of  terms  of  a  decreas- 
ing Geometric  Progression  is ; 

or  more  briefly,  the  sum  to  infinity  is        '■ — . 


THE   SUM   OF  AN   INFINITE  NUMBER   OF   TERMS.     363 


This  quantity,  ,  which  we  call  the  sum  of  the  scries, 

1  —  r 

is  the  limit  to  which  the  sum  approaches,  but  never  actually 

attains;  that  is,  although  no  definite  number  of  terms  will 

amount  to ,  yet  by  taking  a  sufficient  number,  the  sum 

1  —  r 

will  reach  it  as  near  as  we  please. 

1.    Sum  the  series  |,  |,  -J, 

For  n  terms  we  have  by  (2)  of  Art.  163, 


_.*-*) 


1-1. 

2" 


From  this  result  it  appears  that  however  many  terms  be 
taken,  the  sum  of  this  series  is  always  less  than  1.     Also  we 

see  that  by  taking  n  large  enough,  the  fraction  —  can  be 

made  as  small  as  we  please.  Hence  by  taking  a  sufficient 
number  of  terms,  the  sum  can  be  made  to  differ  from  1  by 
as  little  as  we  please  ;  and  when  n  is  made  infinitely  great  we 
have  S  =  1. 

This  may  be  illustrated  geometrically  as  follows: 

A\ 1 1 1 1 \B 

Let  AB  be  a  line  of  unit  length.  Bisect  AB  in  P, ;  bisect  PJ>  in 
P2,  P2B  in  P3,  P3B  in  P4,  and  so  on  indefinitely,  always  bisecting 
the  remaining  distance.  It  is  evident  that  by  a  series  of  such  bisections 
we  can  never  reach  B,  because  we  shall  always  have  a  distance  left 
eqnal  to  half  the  preceding  distance ;  but  by  a  sufficient  number  of 
these  bisections  we  can  come  nearer  to  B  than  any  assigned  distance, 
however  small,  because  every  bisection  carries  us  over  half  the  remain- 
ing distance.     That  is,  if  we  take  a  sufficient  number  of  terms  of  the 


we  shall  have  a  result  differing  from  AB,  i.e.,  from  unity,  by  as  little 
as  we  please.     This  is  simply  a  geometric  way  of  saying  that 


^  +  i  +  ^-  + 
2      2*      2* 


364  VALUE   OF  A    REPEATING    DECIMAL. 

Sum  the  following  series  to  infinity : 

2.  1,  £,1 Arts.  2. 

3.  9,  -G,  4,  - 5f. 

4.  1,  -*»*, |. 

5-  i,i,A, |. 

166.  Value  of  a  Repeating*  Decimal.  —  Repeating 
decimals  furnish  a  good  illustration  of  infinite  Geometric 
Progressions. 

1.  Find  the  value  of  .423. 

.423  =  .4232323  

=  ±  23         23_        

10  103        105 

_  j4_  23 

10  10 


jsV  ^  io2     io4  ^  / 

^  +  ^{77Ll)[,,y(1)ofArt-1G5] 

\  102/ 


=  ±        23   , 


_   4         23        100  =   4  23    =  419 

10        IO3        99         10       990  ~  990' 

which  agrees  with  the  value  found   by  the   usual   rule   in 

Arithmetic. 

The  value  of  any  repeating  decimal  may  be  found  by  the  method 
employed  in  the  last  example;  but  in  practice  it  maybe  found  more 
easily  by  a  general  rule,  which  may  be  proved  as  follows: 

Let  P  denote  the  figures  which  do  not  repeat,  and  suppose  them  p 
in  number;  let  Q  denote  the  repeating  period  consisting  of  (/  figures. 
Let  8  denote  the  value  of  the  repeating  decimal ;  then 

8  =  .PQQQ ; 

.-.    LOpS  =  P.QQQ ; 

and  10p+«<S  =  PQ.QQQ ; 

*  Culled  ulwo  recurring  and  circulating. 


HARMONIC  PROGRESSION.  365 


by  subtracting, 

(10/'+?  —  10p)S  =  PQ  —  P; 
that  is,         10"(10?  -  1).S  =  PQ  —  P; 
.     «_     PQ-P 


(10'/  -  1)10" 

Now  109  —  1  is  a  number  consisting  of  //  nines;  therefore  the 
denominator  consists  of  q  nines  followed  by  p  ciphers.  Hence,  for 
finding  the  value  of  a  repeating  decimal,  we  have  the  following 

Rule.  Subtract  the  integral  number  consisting  of  the  non-repeating 
figures  from  the  integral  number  consisting  of  the  non-repeating  and 
repeating  figures,  and  divide  by  a  number  consisting  of  as  many  nines 
as  there  are  repeating  figures  followed  by  as  many  ciphers  as  there  are 
non-repeating  figures. 

Find  the  value  of  the  following  repeating  decimals  : 


2.    .151515  ....  Arts.  &. 


3.    .123123123  ....    ^fr. 


4.  .16  . 

5.  .037. 


HARMONIC     PROGRESSION. 

167.  Definition.  —  A  series  of  quantities  is  said  to  be  in 
Harmonic  Progression  when  their  reciprocals  are  in  Arith- 
metic Progression. 

Thus,  the  following  series 

hhhh and  J,  f  |,|, 

are  each  in  Harmonic  Progressiou    (II.  P.)   because   their 
reciprocals, 

1,  3,  5,  7,  .....  .  and  4,  3£,  3,  2£, 

are  in  A.  P. 

The  following  is  therefore  a  general  form  for  a  H.  P. 
Ill  1 


a  a  +  d   a  +  2d  a  +  (n  —  l)d 

because  the  reciprocals  of  the  terms  are  in  A.  P. 

From  the  above  definition   it  follows   that   all   problems 
relating  to  quantities  in  H.  P.  can  be  solved  by  taking  the 


366  HARMONIC  MEAN. 

reciprocals  of  the  quantities  and  using  the  formulae  relating 
to  A.  P.  This  makes  it  unnecessary  to  give  any  special 
rules  for  the  solution  of  problems  in  H.  P. 

If  a,  b,  c  be  three  cousecutive  terms  of  a  H.  P.,  then  we 
have  by  definition, 

1  _  1  _  1  _  1 

b       a        c        b 

a  —  b  _  b  —  c 

<ib  be 

.-.     a  —  b  :  b  —  c  :  :  a  :  c       .      .      .      .    (1) 

Thus,  if  three  quantities  are  in  Harmonic,  Progression,  the 
difference  between  the  first  and  the  second  is  to  the  difference 
between  the  second  and  the  third  as  the  first  is  to  the  third. 

Sometimes  this  relation  is  taken  as  the  definition  of  Har- 
monic Progression. 

1.  The  12th  term  of  a  H.  P.  is  ±,  and  the  19th  term  is 
■£%  ;  find  the  series. 

Here  the  12th  and  19th  terms  of  the  corresponding  A.  P. 
are  5  and  2.f  respectively.     Therefore  by 

(1),  Art.  161,  5    =  a  +  He?, 

and  ^f  =  a  +  18d. 

Solving,  we  get  d   =  ^,  a  =  f. 

Hence  the  A.  P.  is  f,  f,  2,  $,  ■£ , 

and  the  II.  P.  is      f,  f,  £,  f ,  f, 

Find  the  last  term  of  the  following  harmonic  series  : 

2.  1,       2,     1-i, to     (!  terms.  Ans.  ■§. 

3.  21,  1{%,  1-&, to  21  terms.  £. 

4.  1$,  lfi,  2T25, to    8  terms.  —4. 

168.  Harmonic  Mean.  —  When  three  quantities  are  in 
Harmonic  Progression,  the  middle  one  is  called  the  Harmonic 
Mean  between  the  other  two. 


HARMONIC  MEAN.  367 

Thus  if  a,  &,  c  arc  in  H.  P.,  6  is  the  harmonic  mean 
between  a  and  c ;  and  by  the  definition  of  H.  P.  we  have 

1  _  1  =  1  _  1. 
6       a        c        b ' 

•••  I-1*- 

b       a       c 
2ac 


a  +  c 


Thus,  ^e  harmonic  mean  between  any  two  quantities  is 
their  product  divided  by  their  sum. 

Between  any  two  given  quantities  any  number  of  terms 
may  be  inserted  so  that  the  whole  series  thus  formed  shall 
be  in  H.  P. ;  the  terms  thus  inserted  are  called  the  harmonic 
means. 

For  example,  to  insert  5  harmonic  means  between  f  and 

A- 

Here  we  have  to  insert  5  arithmetic  means  between  | 
and  Jg5-.     Hence,  by  (1)  of  Art.  161, 

^  =  |  +  Gd;         .-.     d  =  TV 

Thus  the  A.  P.  is  f ,  ff,  f|,  ft,  ff,  ff,  V- ; 
and  therefore  the  required  harmonic  means  between  |  and  ^ 


In  general.     To  insert  n  harmonic  means  between  a  and  b. 
Here  we  have  to  insert  n  arithmetic  means  between  -   and 


a 


-,     By  Art.  162,  these  will  be 
b         J 

11  11 

1,6       a    1    .    0  b       a 

—  -|-  ,  —  -\~  L  i 

a       n  +  \    a  n  +  1 

*wk    (n  +  \)b  +  («  -  b)    (»  +  1)6  +  2 (a  -  6) 
thatlS'  (»  +  l)a6  '  (n  +  \)ab 


368  RELATION   BETWEEN   THE   THREE   MEANS. 

and  therefore  the  required  harmonic  means  between  a  and  b 
are  the  reciprocals  of  these,  that  is, 

(n  +  !)«&        _               (n  +  1)  nb 
(n  +  1)6  +  (a  -  &)'  (n  +  1)6  +  2(a  -  6)' 

1.  Insert  2  harmonic  means  between  4  and  2.     ^4?is.  3,  J^2-. 

2.  Insert  3  harmonic  means  between  ^  and  ^t-     t2s >  tV  3V 

3.  Insert  4  harmonic  means  between  1  and  6.    1 1,  1^,  2,  3. 
169.  Relation  between  Arithmetic,  Geometric,  and 

Harmonic  Means. 

(1)  If  A,  G,  H  be  the  arithmetic,  geometric,  and  har- 
monic means  between  a  and  b,  then  (Arts.  1G2,  164,  168), 

A  =  ^A (1) 

G  =  Va6 (2) 

H=™i- (3) 

a  +  b  K  ' 

Therefore    AH  =  ?LdL&  x  -^-  =  a&  =  £2; 

2  a  +  & 

that  is,  6r  is  the  geometric  mean  between  A  and  //. 

Hence  the  geometric  mean  between  any  two  real  positive 
quantities,  a  and  b,  is  also  the  geometric  mean  between  the 
arithmetic  and  the  harmonic  means  between  a  and  b. 

(2)  From  (1)  and  (2)  we  have 

A  -  G  =  ±±±  -  s/al,  =  Wa  ~  fb)*i 

and  from  (2)  and  (3), 

o-b=^-J^-  =  -%(.^-  VS)'. 

a  +  b       a  -}-  b 
Now  if  a  and  b  arc  both  positive,  \<r  and  V"  are  both  real ; 
therefore  (v'a  —  V^)2  is  positive;  also  \Jab  and  </  +  &  are 
both    positive.     Hence  -4  —  (7   and    Gv  —  //  are    positive. 
Therefore  A>  G  >  II. 


EXAMPLES.  369 

That  is,  the  arithmetic,  geometric,  and  harmonic  means 
between  any  two  real  jjositive  quantities  are  in  descending 
order  of  magnitude.* 

Three  quantities,  a,  b,  c,  are  in  A.  I\,  G.  P.,  or  H.  P., 
according  as 

°^±  =  (\    %  or  %  respectively. 
b  —  c       a     b         c 

The  first  follows  from  the  definition  of  A.  P.  (Art.  161). 
In    the    second,  6(a  —  b)  =  a(b  —  c)  ;         .-.     b2  =  ac. 
See  Art.  164. 

The  third  follows  from  (1)  of  Art.  167. 

Harmonic  properties  are  interesting  chiefly  because  of  their  im- 
portance in  Geometry  and  in  the  Theory  of  Sound.  If  there  be  a 
series  of  strings  of  the  same  substance,  the  lengths  of  which  are 
proportional  to  1,  £,  j,  \,  z,  h  ail(l  if  these  strings  are  stretched  tight 
with  equal  force,  and  any  two  of  them  are  sounded  together,  the  effect 
is  found  to  be  harmonious  to  the  ear. 

Notwithstanding  the  comparative  simplicity  of  the  law  of 
its  formation,  there  is  no  general  formula  for  the  sum  of  any 
number  of  terms  in  harmonic  progression. 

EXAMPLES. 

Find  the  last  term  and  sum  of  the  following  series : 

1.  1,  1.2,  1.4, to  12  terms.  Ans.  3.2,  25.2. 

2.  3£,  1,  -If, to  19  terms.  —41  J,  -361. 

3.  64,  96,  128, to  16  terms.  544,  4864. 

Sum  the  following  series  : 

4.  4,  5±,  6f, to  37  terms.  980f 

5.  -3,  1,5, to  17  terms.  493. 

6.  3a,  a,  —a, to  a  terms.  a2(4  —  a). 

7.  31,  2£,  If, to  n  terms.  \o * 

*  These  two  propositions  were  known  to  the  Greek  geometers. 


370  EXAMPLES. 

8.  1|,  HJ-,  2|$, to  n  terms.        Arts.   "(17  +  7"\ 

9.  ■— - ,  V^,  — — ^ — ,  ....  to  7  terms.         7(^2  +  2). 

\ft  +  1  ^2-1 

Find  the  scries  in  which 

10.  The  15th  term  is  25,  and  the  29th  term  46. 

Arts.  4,  ;H,  7, 

11.  The  15th  term  is  -25,  and  the  23d  term  -41. 

Ans.  3,  1,  -1, 

12.  Insert  14  arithmetic  means  between  —  7£  and  —  2i. 

Ans.  -OH,  -6&, -2TV 

13.  Insert  36  arithmetic  means  between  8|  and  2£. 


Ans.  &|,  8$, 


2|. 

How  many  terms  must  be  taken  of  the  series 

14.  15|,  15±,  15, to  make  129?     .4ns.  9,  or  86. 

15.  -10£,  -9,  -lh,  ...  to  make  -42?  7,  or  8. 

16.  -6f,  -6f,  -6,  ....  to  make  -52$ ?  11,  or  24. 

17.  The  sum  of  three  numbers  in  A.  P.  is  33,  and  their 
product  is  792  :  find  them.  Ans.  4,  11,  18. 

18.  An  A.  P.  consists  of  21  terms  ;  the  sum  of  the  three 
terms  in  the  middle  is  129,  and  of  the  last  three  is  237  :  find 
the  series.  Ans.  3,  7,  11, 83. 

19.  The  first  term  of  an  A.  P.  is  5,  and  the  fifth  term  is 
11 :  find  the  sum  of  8  terms.  Ans.  82. 

20.  The  sum  of  four  terms  in  A.  P.  is  H,  and  the  last 
term  is  17:  find  the  terms.  Ans.  5,  9,  13,  17. 

21.  The  seventh  term  of  an  A.  P.  is  12,  and  the  twelfth 
term  is  7  ;  the  sum  of  the  series  is  171  :  find  the  number  of 
terms.  Ans.  bs,  or  L9. 

22.  A  sets  ont  from  a  place  and  travels  2.1  miles  an  hour. 
B  sets  out  3  hours  after  A,  and  travels  in  the  same  direction, 

3  miles  the  first  hour,  .">j,-  miles  the  second.    1   miles  the   third. 
and  so  on.    In  how  many  hours  will  B  overtake  A?     Ans.  •"». 

23.  In  the  series,  1,  8,  5,  etc.,  the  sum  of  2n  terms  :  the 
sum  of  n  terms  :  :  x  :   1  :   find  the  value  of  SB.  Ans.  4. 


EXAMPLES.  371 

24.  Find  an  A.  P.  such  that  the  sum  of  the  first  five  terms 
is  one-fourth  the  sum  of  the  following  five  terms,  the  first 
term  being  unity.  Ans.  1,  —  2,  —5, — 2G. 

25.  If  the  sum  of  m  terms  of  an  A.  P.  be  always  to  the 
sum  of  n  terms  in  the  ratio  of  m2  to  n\  and  the  first  term  be 
unity,  find  the  nih  term.  Ans.  2n  —  1. 

26.  If  2n  +  1  terms  of  the  series  1,  3,  5,  7,  9,  ...  .  be 
taken,  show  that  the  sum  of  the  alternate  terms,  1,  5,  9, 

will  be  to  the  sum  of  the  remaining  terms  3,  7,  11, 

as  n  +  1  to  n. 

27.  On  the  ground  are  placed  n  stones ;  the  distance 
between  the  first  and  second  is  one  yard,  between  the 
second  and  third  three  yards,  between  the  third  and  fourth 
five  yards,  and  so  on  :  how  far  will  a  person  have  to  travel 
who  shall  bring  them  one  by  one  to  a  basket  placed  at  the 
first  stone?  *  An^  n^  _  l){2n  _  1}  y.mls> 

o 

28.  Find  a  series  of  arithmetic  means  between  1  and  21, 
so  that  their  sum  has  to  the  sum  of  the  two  greatest  of  them 
the  ratio  of  11  to  4.      Ans.  9  means,  3,  5,  7, 19. 

29.  Find  the  number  of  arithmetic  means  between  1  and 
19  when  the  second  mean  is  to  the  last  as  1  to  G.      Ans.  17. 

30.  If  the  second  term  of  an  A.  P.  be  a  mean  proportional 
between  the  first  and  the  fourth,  show  that  the  sixth  term 
will  be  a  mean  proportional  between  the  fourth  and  the 
ninth. 

Find  the  last  term  of  each  of  the  following  geometric 
series : 

31.  2,  —6,    18,     to  8  terms.        Ans.  —4374. 

32.  2,      3,      4§ to  G  terms.                          Vr- 

33.  3,  —3'-,    33, to  2n  terms.  —3-". 

Sum  the  following  series  : 

34.  1,  -.V.  \, to  12  terms.  Ans.  i|||. 

35.  cj,  _c,  4, to    7  terms.  5ff. 

*  See  Art.  170. 


372  EXAMPLES. 

36.  2,  —4,  8, to    2;>  terras.    Ans.  |(1  —  22p). 

37.  ^2,  Vg,  3^2, to  12  terms.       364(^6  +  ^2). 

38.  Insert  3  geometric  means  between  486  and  G. 

Ans.    162,  54,  18. 

39.  Insert  4  geometric  means  between  |  and  128. 

Ans.  £,  2,  8,  32. 

40.  Insert  3  geometric  means  between  1  and  256. 

Ans.  4,  16,  64. 

41.  Insert  4  geometric  means  between  3  and  —  729. 

Ans.  -9,  27,  -81,  243. 
Sum  the  following  series  : 

42.  §,      J,       £, to  6  terms.  Ans.  fff . 

43.  1,  — ^,       £, to  infinity.  f. 

44.  6,  —2,       f, to  infinity.  4£. 

45.  },      |,     ^\, to  infinity.  1. 

46.  |,-1,       f, to  infinity.  §f 

47.  .9,  .03,  .001, to  infinity.  §£. 

Find  the  value  of  the  following  repeating  decimals : 

48.  .4282828  .  .  .   Ans.  §£§.  I      50.    .16.  Ans.  £. 

49.  .28131313...         ^ r.  I      51.    .378.  |f. 

52.  The  sum  of  three  terms  in  G.  P.  is  63,  and  the 
difference  of  the  first  and  third  terms  is  45  :  find  the  terms. 

Ans.  3,  12,  48  ;  or  36,  -54,  SI. 

Let  a,  ar,  «r2  denote  the  numbers. 

53.  The  sum  of  the  first  four  terms  of  a  G.  P.  is  40,  and 
the  sum  of  the  first  eight  terras  is  3280 :  find  the  series. 

Ans.  1,  3,  9, 

54.  The  sum  of  three  terms  in  G.  P.  is  21,  and  the  sura 
of  their  squares  is  189  :  find  the  terms.  Ans.  3,  6,  12. 

55.  A  person  who  saved  every  year  half  as  much  again  as 
he  saved  the  previous  year  had  in  seven  years  saved  $102.95  : 
how  much  did  he  save  the  first  year?  Ans,  $3.20. 


EXAMPLES.  373 

5G.  In  a  G.  P.  show  (1)  that  the  product  of  any  two  terms 
equidistant  from  a  given  term  is  always  the  same,  and  (2) 
that  if  each  term  be  subtracted  from  the  succeeding,  the 
successive  differences  are  also  in  G.  P. 

57.  Show  that  the  square  of  the  arithmetic  mean  of  two 
quantities  is  equal  to  the  arithmetic  mean  of  the  arithmetic 
and  geometric  means  of  the  squares  of  the  same  two  quanti- 
ties. 

58.  There  are  four  numbers,  the  first  three  of  which  are 
in  G.  P.,  and  the  last  three  in  A.  P.  ;  the  sum  of  the  first 
and  last  is  14,  and  the  sum  of  the  second  and  third  is  12  : 
find  the  numbers.  Ans.  2,  4,  8,  12. 

59.  Three  numbers  whose  sum  is  15  are  in  A.  P.  ;  if  1, 
4,  and  19  be  added  to  them  respectively,  the  results  are  in 
G.  P.  :  find  the  numbers.  Ans.  2,  5,  8. 

Find  the  last  term  of  the  following  harmonic  series  : 

60.  6,  3,  2,     to  G  terms.  Ans.  1. 

Gl.    8,  2,  li to  6  terms.  J. 

62.    Insert  three  harmonic  means  between  2|  and  12. 

Ans.  3,  4,  G. 

G3.  The  arithmetic  mean  of  two  numbers  is  9,  and  the 
harmonic  mean  is  8  :  find  the  numbers.  Ans.  6,  12. 

G4.  Find  two  numbers  such  that  the  sum  of  their  arith- 
metic, geometric,  and  harmonic  means  is  9|,  and  the  product 
of  these  means  is  27.  Ans.  1,  9. 

65.  The  arithmetic  mean  of  two  numbers  is  3,  and  the 
harmonic  mean  is  -| :  find  the  numbers.  Ans.  2,  4. 

66.  There  are  three  numbers  in  II.  P.,  such  that  the 
greatest  is  the  product  of  the  other  two,  and  if  one  be  added 
to  each,  the  greatest  becomes  the  sum  of  the  other  two : 
find  the  numbers.  Ans.  2,  3,  6. 

67.  The  sum  of  two  consecutive  terms  in  H.  P.  is  T2n9T, 


374  MATHEMATICAL   INDUCTION. 


CHAPTER     XVIII. 

MATHEMATICAL  INDUCTION  —  CONTINUED 
FRACTIONS  — PERMUTATIONS  AND  COMBI- 
NATIONS. 

170.  Mathematical  Induction  —  General  Proofs  of 
Theorems  Stated  in  Art.  51.  —  Many  mathematical  form- 
ula? are  not  easily  demonstrated  by  a  direct  mode  of  proof ; 
in  such  cases  it  is  often  found  convenient  to  employ  a  method 
of  proof  which  is  known  as  MatJiematiccU  Induction.  This 
method  may  be  illustrated  as  follows  : 

1.  Suppose  it  is  required  to  establish  the  truth  of  the 
following  formula : 

p  +  2»  +  32  + +  n2  =  »(tt  +  l)(g»+l). 

G 

We  can  easily  see  hy  trial  that  this  formula  is  true  in 
simple  cases,  such  as  when  n  =  1,  or  2,  or  3  ;  and  from  this 
we  might  be  led  to  conjecture  that  the  formula  was  true  in  all 
cases. 

Assume  that  it  is  true  for  n  terms.  Add  (n  +  l)a  to  both 
members ;  then 

!2+2a+3a+.  .  .+wa+(yi+1)2=n(n+l)(2n+l)  +  (w+1)a 

G 

=  (n  +  l^±J)+u  +  1j 

=   (ft  +  l)(u  +  2)('2n  +  3)  . 
G 

which  is  the  same  formula  for  the  sum  of  n  +  1  terms  of  Ihe 
series  that  we  assume  to  be  true  for  n  terms,  n  +  1  taking 
the  place  of  n.     In  other  words,  if  the  formula  is  true  when 


MATHEMATICAL   INDUCTION.  375 

we  take  a  certain  number  of  terms,  whatever  that  number 
may  be,  it  is  true  when  we  increase  that  number  by  one. 
But  we  see  that  it  is  true  when  3  terms  are  taken  ;  therefore 
it  is  true  when  4  terms  are  taken  ;  and  therefore  it  is  true 
when  5  terms  are  taken ;  and  so  on.  Hence  the  formula  is 
true  universally. 

2.  The  first  theorem  stated  in  Art.  51  may  be  proved  by 
this  method  as  follows  :  By  division 

a"  -  t  __  xn-i  +  y(af-1-yB-1). 
x  —  y  x  —  y 

hence  if  x"~l  —  yn~l  is  divisible  by  x  —  y,  then  xn  —  yn  is 
divisible  by  x  —  y.  But  we  know  that  x2  —  y2  is  divisible 
by  a;  —  y ;  therefore  Xs  —  y3  is  divisible  by  x  —  y  ;  and 
therefore  x*  —  yi  is  divisible  by  x  —  y ;  and  so  on  indefi- 
nitely. Hence  xn  —  yn  is  always  divisible  by  x  —  y,  when  n 
is  any  positive  integer. 

Note.  —  The  two  theorems  which  we  have  here  j)roved  by  the 
method  of  induction  may  be  established  by  other  methods.  There  are 
many  other  theorems  which  are  capable  of  easy  proof  by  the  method 
of  induction;  and  in  the  subsequent  part  of  this  book  we  shall  some- 
times have  occasion  to  use  this  method.  Tbe  student  of  Natural 
Philosophy  will  find  the  word  induction  used  in  a  different  sense  in 
that  subject  from  what  it  has  here.  There  we  cannot  be  sure  that 
the  law  holds  in  any  cases  except  those  which  we  have  examined.  In 
fact,  induction,  as  used  in  Natural  Philosophy,  is  never  absolutely 
demonstrative,  and  often  far  from  it;  whereas  the  method  of  mathe- 
matical induction  is  as  rigid  as  any  other  process  in  mathematics. 

If  we  change  y  into  —y,x—y  becomes  x  —  (—y)  =  x  +  y; 
also  x"  —  yn  becomes  xn  —  (  —  y)n,  which  is  xn  —  y*  or  .<;"  +  y", 
according  as  n  is  even  or  odd. 

Hence,  when  n  is  even,  xn  —  yn  is  divisible  b}^  x  -\-  y  ;  and 
when  n  is  odd,  x11  +  y"  is  divisible  by  x  +  y. 

Since  af*  -f-  yn  =  xn  —  yn  +  2yn,  it  follows  that  when 
xn  +  y"  is  divided  by  x  —  ?/,  the  remainder  is  2y" ;  therefore 
xn  +  yn  is  never  divisible  by  x  —  y. 


376  CONTINUED   FRACTIONS. 


CONTINUED     FRACTIONS. 

171.  Definition.  —  A  Continued  Fraction  is  one  whose 
denominator  is  a  whole  number  plus  a  fraction  ;  the  denomi- 
nator of  this  last  fraction  a  whole  number  plus  a  fraction, 
and  so  on.     Thus, 

a  + b- a  + l- 


c  + =—  ,  b  + 


f 
—J——  c  + 


g  +  etc.  d  +  etc. 

are  continued  fractions;  but  the  term  is  commonly  restricted 
to  the  latter  form,  where  all  the  numerators  of  the  partial 

fractions  are  1,  and  all  the  denominators,  a,  b,  c, are 

positive  integers. 

For  economy  of   space    the   continued    fraction    is   often 
written  in  the  more  compact  form 


a  + 


b+   c+   d  -f  etc. 


A  continued  fraction    may  either  terminate  with   one    of 
the  denominators,  or  it  may  extend  indefinitely.     When  the 

number  of   terms,  a,  b,  c, is  finite,  the   fraction  is 

said  to  be  terminating ;  if  the  number  of  terms  is  unlimited, 
the  fraction  is  called  an  infinite  continued  fraction.  A  ter- 
minating continued  fraction  may  be  reduced  to  an  ordinary 
fraction  by  simplifying  the  fractions  in  succession,  beginning 
with  the  lowest. 

172.  To  Convert  a  Given  Fraction  into  a  Continued 
Fraction. 

Let  —  be  the  oiven  fraction  ;  divide  m  by  n.  let  a   be  the 

n  J 

quotient  and  p  the  remainder  ;  thus 


m 

= 

a 

+ 

P 

_ 

a 

+ 

1 

n 

n 

n 

CONTINUED   FRACTIONS.  377 

Divide  n  byj9,  let  6  be  the  quotient  and  q  the  remainder; 
thus 

5=  &  +  2=  6  +  l. 

2>  V  P 

fl 

Divide  p  by  q,  let  c  be  the  quotient  and  r  the  remainder  ;  and 
so  on.     Thus 

m  ,  1  ,11 

—  =  «  -i =  a  ^ — ■ -. 

n  6  4-         1  6+  c  +  etc. 

c  +  etc. 

If  m  <  n.,  the  first  quotient  a  is  zero. 

It  will  be  seen  that  the  above  process  is  the  same  as  that 
of  finding  the  greatest  common  divisor  of  m  and  n.  Hence 
we  shall  at  last  arrive  at  a  point  where  the  remainder  is  zero, 
and  the  process  terminates.  Thus  every  fraction  whose 
numerator  and  denominator  are  positive  integers  can  be 
converted  into  a  terminating  continued  fraction. 

The  fractions  formed  by  taking  one,  two,  three,  etc.,  of 
the  quotients  of  a  continued  fraction  are  called  the  first, 
second,  third,  etc.,  converging  fractions,  or  convergents, 
because,  as  will  be  shown  in  Art.  176,  each  successive  con- 
vergent is  a  nearer  approximation  to  the  true  value  of  the 
fraction  than  any  of  the  preceding  convergents.  Thus,  the 
first  convergent  is  a ;  the  second  convergent  is  formed  from 

«  H — ,  which  is  therefore  — —  ;   the   third  convergent  is 

b  b 

formed  from  ft  ^ ,  which  is  therefore  a  + 


6+i  bc+l 

c 

abc  +  «  +  c         t 

— ^—  ;  and  so  on. 

6c  +  1 

173.  The  Successive  Convergents  are  Alternately 
Less  and  Greater  than  the  Continued  Fraction. — 

The   first   convergent,    ft,    is   too    small   because    the    part 


378  SUCCESSIVE   CONVERGENTS. 


is  omitted.     The  second  convergent,  a  -\ — ,  is  too 


b  +  etc.  6' 

great,  because  the  denominator  b  is  too  small.     The  third 

convergent,  a  +  ,  is  too  small  because  6  +  -  is  too 

c 
great ;  and  so  on. 

When  the  given  fraction  is  a  proper  fraction,  a  =  0 ;  if 
in  this  case  zero  be  considered  as  the  first  convergent,  the 
above  results  may  be  enunciated  as  follows  : 

The  convergents  of  an  odd  order  are  all  less,  and  the  con- 
vergents  of  an  even  order  are  all  greater,  than  the  continued 
fraction. 

174.  To  Prove  the  Law  of  Formation  of  the  Suc- 
cessive Convergents.  —  The  first  three  convergents  are 
(Art.  172) 

a     ab  +  1     c(ab  +  1)  +  a . 
l'         b      '  be  +  1 

we  see  that  the  numerator  of  the  third  convergent  may  bo 
formed  by  multiplying  the  numerator  of  the  second  conver- 
gent by  the  third  quotient,  and  adding  the  numerator  of  the 
first  convergent ;  also  that  the  denominator  of  the  third 
convergent  may  be  formed  in  a  similar  manner  by  multiply- 
ing the  denominator  of  the  second  by  the  third  quotient,  and 
adding  the  denominator  of  the  first.  We  shall  now  prove  by 
induction  that  this  law  holds  universally. 

Let  the  numerators  of  the  successive  convergents  be 
denoted  by  pv  p2,  jp8,  etc.,  and  the  denominators  by  qv  q.„  qs, 
etc.;  oa,  a2,  a3,  etc.,  the  corresponding  quotients;  and 
assume  that  the  law  of  formation  above  stated  holds  for  the 
u"'  convergent,  so  that 

Since  the  (n  +  1)"'  convergent  differs  from  the  »tt  only 


DIFFERENCE  BETWEEN   CONVERGENTS.  379 


iii  having  the  quotient  a„  -] instead  of  a„,  it  follows  that 


this  convergent,  or 

<2n  +  l 


+  JV 


<»  +  —  YJn-i  +  qn->. 


+  l(«»P.-l    +Pn-2)    +P,-1 

+iK&-i  +  g„_2)  +  <?»_i 

+  lft    +  J>n-1 


«»+ig»  +  g„_i 

Hence  if  we  put pB+i=a»  +  ii\+P»-n  g»  +  i  =  «»  +  i7»+f7»-n 
we  see  that  the  (n  +  l)th  convergent  follows  the  same  law 
that  was  supposed  to  hold  for  the  nth  convergent.  But  the 
law  does  hold  for  the  third  convergent ;  therefore  it  holds 
for  the  fourth,  and  so  on  ;  therefore  it  holds  universally. 

Reduce  ff|  to  a  continued  fraction,  and  find  the  successive 
convergents. 

By  the  process  of  finding  the  greatest  common  divisor  of 
323  and  117,  the  successive  quotients  are  found  to  be  2,  1, 
3,  5,  1,  1,  2.     Thus  (Art.  172) 

m  =  2  +  — - — T 


3  + 


1  + 


i+i 


or  2  H . 

1+  3+  5+  1+  1+2 

The  successive  quotients  are  2,  1,  3,  5,  1,  1,  2. 

The  successive  convergents  are  f,  f ,  J^1-,  ff,  ff,  -^y,  fff . 

175.  Difference  Between  Two  Consecutive  Con- 
vergents. —  The  difference  between  any  two  consecutive  con- 
vergents is  a  fraction  whose  numerator  is  unity,  and  whose 


380    LIMIT   OF  ERROR   IN   TAKING  ANY  CONVERGENT. 

denominator  is  the  product  of  the  denominators  of  the  con- 
venient*. 

This  is  evident  with  respect  to  the  first  two  convergents  ; 

~         A    ,    .  „ ,    ab  +  1        a        1 

for,  Art.  1/4, ±— =  -. 

b  16 

Assume  the  law  to  hold  for  any  two  consecutive  conver- 
gents, 2- ,  i-±  ;  so  that 

Pi  _  P  __  PiQ  ~  PQi*  =  J_  . 

(h  ~  3  ^i  99j '      m     '     '  ^  > 

then,  a,  ax,  o2,  etc.,  being  the  corresponding  quotients,  we 
have  (Art.  174) 

Pz  _    Pi  _    P29l    ~    Plg2 
?2   ~     ft  ^1^2 

_  ("2fli  +  P)ffi  ~  K9i  +  g)pt 

=  ^T-  =  ^'by(1)' 

Hence  if  the  law  holds  for  one  pair  of  consecutive  conver- 
gents, it  holds  for  the  next  pair.  But  the  law  does  hold  for 
the  first  pair,  therefore  it  holds  for  the  second  pair,  and  so 
on.     Hence  it  is  universally  true. 

Note.  —  In  computing  the  numerical  value  of  the  successive 
convergents,  this  theorem  furnishes  an  easy  test  of  the  accuracy 
of  the  work. 

All  convergents  are  in  their  lowest  terms;  for  if  p  and  q 
had  a  common  divisor  it  would  divide  })xq  ~  pqv  or  unity, 
which  is  impossible. 

176.  Limit  of  Error  in  Taking  any  Convergent.  — 
(1)  Every  convergent  is  nearer  to  the  continued  fraction  than 
an y  of  the  preceding  convergents. 

Let  x  denote  the  continued  fraction,  and  —  ■>  —  i  — ,  three 

9    Qi    & 

•  The  sign   ^  means  "  the  difference  between." 


LIMIT   OF  ERROR   IN   TAKING  ANY   CONVERGENT.    381 


p., 
consecutive   convergents;    then   x  differs    from   —   only  in 

72        J 

taking  the  complete  quotient  «.,  +  instead  of  a2. 

a3  +  etc. 

Denote  this  complete  quotient,  which  is  always  greater  than 

unity,  by  k  ;  thus 


_  fc?>i  +  P  . 
kq1  +  q  ' 


...     x„P  =  Hf>~P^)=         *        .[Art.  175,(1)]    (1) 

q         q{kqx  +  q)         q(kqx+  q)' '  L  WJ     W 

andB^a;=   frg~.Wi    = I . 

9i  <ii(Mi  +  v)     qAMi  +  q) 

Now  A;  >  1,  and  q  <  qx ;  hence  for  both  these  reasons 

Pi  P . 
'  x  <  x  ~  -  ) 

9i  q 

Pi .  p 

that  is,  —  is  nearer  to  x  than  —  is.     Hence  every  convergent 

is  nearer  to  the  continued  fraction  than  the  next  preceding 
convergent  is,  and  therefore  than  any  preceding  convergent. 

Comparing  this  result  with  that  of  Art.  173,  we  see  that 

The  convergents  of  an  odd  order  continually  increase,  but 
are  always  less  than  the  continued  fraction. 

The  convergents  of  an  even  order  continually  decrease,  but 
are  always  greater  than  the  continued  fraction. 

(2)  To  find  limits  to  the  error  made  in  taking  any  conver- 
gent for  the  continued  fraction. 

Since  k  >  1,  we  see  from  (1)  that  the  difference  between  x 

and|  <i-,and> 1 (2) 

^  Wi  7(7!  +  7)  K  J 

p 
Also,  since  ql  >  q,  the  error  in  taking  —  instead  of  x 

<  - - ,  and  >  — . 
q-  2ft2 


382    LIMIT   OF  ERROR   IN    TAKING  ANY  CONVERGENT. 


Therefore,  as  the  error  is  less  than  — ,  it  follows  that  in 

r 

order  to  find  a  convergent  which  will  differ  from  the  con- 
tinued fraction  by  less   than   a  given   quantity  -,  we  have 

a 

only  to  form  the  consecutive  convergeuts  until  we  reach  one 

P 

— ,  where  cr  is  not  less  than  a. 

(3)    It   appears    from  (2)  that  the  error  in   taking   any 

Vn 

convergent,  — ,  instead  of  the  continued  fraction,  x 
< ,  or  — (Art.  174) ;  that  is,  < 


QnQn  +  l  Qni'tn  +  Sln    +    ^-j)  «„  +  ^ 

Pn 

Hence  the  larger  an  +  1  is,  the  nearer  does  —  approximate 

On 

to  the  continued  fraction  ;  therefore 

Any  convergent  which  immediately  precedes  a  large  quotient 
is  a  near  approximation  to  the  continued  fraction. 

(4)   Any  convergent  is  nearer  to  the  continued  /ruction  titan 

any  other  fraction  whose  denominator  is  less  than  that  of  the 

convergent. 

p   p. 
Let  — ,  —  be  two  consecutive  convergeuts,  x  the  continued 
Q    Q\ 

fraction,  and  -  a  fraction  whose  denominator  s  is  less  than  qv 

r  p>\     ,         r 

If  possible,  let   -  be  nearer  to  x  than  —   :  then  -   must  be 
1  *  <h  s 

V  P 

nearer  to  x  than      ,  l>v  ( 1 )  ;    and  .since  x  lies  between       and 

V\  r  V  V\ 

— ,  it  follows  that  -  must  lie  between       and  — .     Hence 

7i  s  g  7l 

r        p     p,       p  1 

.  ~  £<* -.that  is  <-      (Ait.  17;")). 


rq  ~.  sp 


APPROXIMATE  VALUE    OE   A    QUADRATIC  SURD.      383 

that  is,   an  integer  less  than   a   proper  fraction,   which  is 

Pi 

impossible.     Therefore  —  must  be  nearer  to  the  continued 
Sri 

fraction  than  -  is. 

6' 

Find  a  series  of  fractions  approximating  to  3.14159. 

By  the  process  of  finding  the  greatest  common  divisor  of 
314159  and  100000,  the  successive  quotients  are  3,  7,  15,  1, 
25,  1,  7,  4.     Thus 

3.14159  =  3  +JL_J_J_J_J_J_1. 

7+  15+  1+  25+  1+  7+  4 

The  successive  convergents  are  f ,  -2T3-,  f$f,  fff , ; 

this  last  convergent  which  precedes  the  large  quotient  25 
is   a  very  near  approximation,   the   error  being   less   than 

- ,  by  (3),  and  therefore  less  than  .000004. 

25(113)*      J  V   " 

The  method  of  continued  fractions  thus  enables  us  to  find 
two  small  integers  whose  ratio  closely  approximates  to  that 
of  two  quantities  whose  exact  ratio  can  only  be  expressed 
by  large  integers. 

177.  Approximate  Value  of  a  Quadratic  Surd. — 
By  the  properties  of  continued  fractions  we  may  approximate 
to  the  value  of  a  quadratic  siird.     Thus, 

Express  Vll  as  a  contiuued  fraction,  and  find  a  series  of 
fractious  approximating  to  its  value. 


vTT  =  3  +  (vTl  -  3)  =  3  + 


2 


\/TT 


V41  +  3        -         Vll   -  3  „  1 

o —  '->  +    o — —  —  o  -\ — — , 

2  2  V/ll  +  3 

V/TT  +  3  =  G+    ^11-3  =6  +  v/^^5 

after  this  the  quotients  3,  6,  will  be  repeated  indefinitely 

hence  1111 

Vll  =  3  +     ' 


3+  6+  3+  6  +  etc. 


384  PERIODIC   CONTINUED   FRACTIONS. 

Explanation. —In  each  line  of  the  above  process  we  perform 
the  same  series  of  operations.  Thus,  in  the  first  line  we  find  the 
greatest  integer  in  ^11,  which  is  3,  and  the  remainder  is  <J\\  —  3. 
We  then  rationalize  the  numerator,  so  that  after  inverting  the  result 

,  we  begin  a  new  line  with  a  rational  denominator. 

vTT  +  3 

The  quadratic  surd  is  thus  expressed  in  the  form  of  an 
infinite  continued  fraction  (Art.  171).  It  will  be  noticed 
that  the  quotients  begin  to  recur  as  soon  as  we  arrive  at  a 
quotient  which  is  double  of  the  first.  It  may  be  proved  that 
this  is  always  the  case  ;  *  but  the  proof  is  too  long  and 
tedious  for  this  work.  Airy  particular  example  can  be  solved 
like  the  above. 

Since  the  quotients  are  3,  3,  6,  3,  G,  etc.,  the  first  five 
convergents,  formed  by  the  rule  in  Art.  174,  are  f,  Y',  if' 

13JL      1_25_7 
tfo  i     379  ' 

The  error  in  taking  the  last  of  these  is  less  than 


(379)2 
178.   Periodic  Continued  Fractions.  —  A  continued 

fraction   in  which   the   denominators   repeat   themselves   is 

called  a  periodic  continued  fraction.     Thus,  the  continued 

fraction  in  Art.  177  is  periodic. 

A  periodic  continued  fraction  is  equal  to  one  of  the  roots 

of  a  quadratic  equation. 

Let  x  denote  the  value  of  the  continued  fraction  - —  - — 

3+  6-f- 

— - ;  then 

3+  G  +  etc. 

1  .!„.  -.    ,.         6  +  x 


x  = 


that  is,  x  = 


3  +  J-  l8  +  S* 

6  +x 


whence  sc2  +  6x  =  2,  a  quadratic  equation. 

The  continued  fraction  is  equal  to  the  positive  root  of 
this  equation,  and  is  therefore  equal  to  —3  +  VI 1  -  (See 
example  in  Art.  177.) 

*  Todhuntor'e  Algebra,  p.  376;  also  Ball  and  Knight,  i>.  -J97. 


EXAMPLES.  385 

EXAMPLES. 

Reduce  each  of  the  following  fractions  to  a  continued 
fraction,  and  find  the  fourth  convergent  to  each. 

1.    253.  ^S.l+J_J_J_JL_LJ_JL;L7. 

179  2+2+2+1+1+2+2+12 

2  1189  1111111.    33 

3927'  3+  3+  3+  3+  3+3+3'  109* 

3  37  J_  JL  J_  J_  _L  I  •  _Z_ 

2+  1+  2+  2+1+3'  19' 

1    84  1       1       1       1       1       11.7 

'227'  2+  1+  2+  2+  1+  3+  2'   19' 

5.  Express  y/l8  as  a  continued  fraction,  and  find  the  limits 
of  the  error  made  by  taking  the  fourth  convergent. 

Ans.    4  -\ ;  — ,  and  the   error  lies 

4+  8+  4+  8+ etc.     136 

between  — and (Art.  176,  (2)). 

136x1121  136  x  1257  " 

PERMUTATIONS    AND    COMBINATIONS. 

179.  Definitions.  —  The  different  orders  in  which  a 
number  of  things  can  be  arranged,  either  by  taking  some  or 
all  of  them,  are  called  their  Permutations. 

Thus,  the  permutations  of  the  letters  a,  b,  c,  taken  one  at 
a  time  are  three,  viz.,  a,  b,  c;  taken  two  a  time,  are  six, 
viz.,  ab,  ba,  ac,  ca,  be,  cb ;  and  taken  three  at  a  time,  are 
also  six,  viz., 

abc,  acb,  bca,  bac,  cab,  cba. 

The  Combinations  of  things  are  the  different  groups  or 
collections  which  can  be  made,  either  by  taking  a  part  or  all 
of  them,  without  reference  to  the  order  in  which  the  things 
are  placed. 

Thus,  the  combinations  of  the  letters  a,  b,  c,  taken  two  at 
a  time  are  three,  viz.,  ab,  ac,  be;  ab  and  ba,  though  differ- 
ent permutations,  form  the  same  combination,  both  consisting 
simply  of  a  and  b  grouped  together. 


386  THE   NUMBER    OF  PERMUTATIONS. 

It  appears  from  this  that  in  forming  combinations  we  are 
concerned  only  with  the  number  of  things  eacli  group  con- 
tains ;  while  in  forming  permutations  we  have  also  to  consider 
the  order  of  the  things  which  make  up  each  group  ;  thus  the 
above  six  permutations  of  the  letters  a,  b,  c,  taken  three  at 
a  time,  form  but  one  combination. 

180.  The  Number  of  Permutations. —  To  find  the 
number  of  permutations  of  n  different  things,  taken  r  at  a 
time. 

Let  the  different  things  be  represented  by  n  letters,  a,  b, 

c, Set  a  aside  ;  write  down  the  other  n  —  1  letters 

in  a  line  ;  put  a  before  each  of  them  in  succession  ;  we  thus 
obtain  ab,  ac,  ad,  etc.,  or  n  —  1  permutations,  each  of  two 
letters  in  which  a  stands  first.  In  the  same  manner  there 
are  n  —  1  permutations,  each  of  two  letters  in  which  b 
stands  first.  Similarly  there  are  n  —  1  permutations, 
each  of  two  letters  in  which  c  stands  first ;  and  so  on  for  each 
of  the  other  letters  ;  and  as  there  are  n  of  them,  the  whole 
number  of  permutations  of  the  n  letters,  two  together,  is 
n(n  —  1). 

Again,  set  a  aside,  and  group  the  other  n  —  1  letters,  two 
and  two  ;  as  has  just  been  shown,  there  are  (n  —  1)  (u  —  2) 
such  groups.  Put  a  before  each  of  them,  and  we  have 
(n  —  l)(n  —  2)  permutations,  each  of  three  letters  in  which 
a  stands  first.  Similarly  there  are  (n  —  l)(?i  —  2)  per- 
mutations, each  of  three  letters  in  which  b  stands  first ;  and 
so  on  for  each  of  the  other  letters.  Therefore  the  whole 
number  of  permutations  of  n  letters  taken  three  at  a  time  is 
n(n  -  l)(w  -  2). 

Proceeding  thus,  and  noticing  that  at  any  stage,  the 
number  of  factors  is  the  same  as  the  number  of  letters  in 
each  permutation,  and  that  the  negative  number  in  the  Inst 
factor  is  one  less  than  the  number  of  letters  in  each  permu- 
tation, we  shall  have  the  number  of  permutations  of  n  things, 

r  together  equal  to  n(n  —  1 )  (n  —  2) to  r  factors  ; 

and  the  rth  factor  is  n  —  (r  —  1)  or  n  —  r  +  1. 


THE  NUMBER   OF  PERMUTATIONS.  387 

Hence,  the  whole  number  of  permutations  of  n  things  taken 
r  at  a  time  is 

n(n  -  1)  (n  -  2) (n  -  r  +  1)  .     .   (1) 

If  all  the  letters  are  taken  together,  r  =  n,  and  (1) 
becomes 

n(n  -  l)(n  -  2) 3-2-1   .     .     .   (2) 

Hence,  the  member  of  permutations  of  n  things  taken  all  at 
a  time  is  equal  to  the  product  of  the  natural  numbers  from  1 
up  to  n. 

It  is  usual  to  denote  this  product  by  the  symbol  [n,  which 
is  read  "  factorial  n."  *     Thus, 

Factorial  6,  or  [6,  means  6  •  5  •  4  •  3  •  2  •  1,  or  720  ; 
factorial  5,  or  15,  means  5  •  4  •  3  •  2  •  1,  or  120. 

From  the  law  of  formation  it  is  clear  that  17  =  716. 

More  generally,  \n  -\-  1  =  (n  +  l)[w. 

Thus  \n  +  1  contains  all  the  factors  of  [n,  and  one  factor, 
n  +  1?  additional. 

Denoting  the  number  of  permutations  of  n  things  taken  r 
at  a  time  by  the  symbol  „Pr,  we  have  from  (1)  and  (2) 

HPr=n(n— l)(w— 2).  .  .  .(»— r+1);     and     nPn=\n. 

Thus,  nP4=w(n-l)0-2)0-3). 

Also    „P5=nP4(?i-4)  =n(w-l)  (n-2)  (»-3)  (n-4)  ; 

and  so  on. 

1.  Four  persons  enter  a  railway  carriage  in  which  there 
are  six  seats ;  in  how  many  ways  can  they  take  their  places? 

Here  n  =  6,  and  r  =  4  ;  then  by  (1)  we  have 

6P4  =  6  •  5  •  4  •  3  =  360. 

2.  Required  the  number  of  changes  which  can  be  rung, 
(1)  with  5  bells  out  of  8,  and  (2)  with  the  whole  peal. 

Ans.  (1)  6720;   (2)  40320. 

*  It  in  also  sometimes  denoted  by  n\. 


388  THE  NUMBER   OF   COMBINATIONS. 

3.  Required  the  number  of  different  ways  in  which  6 
persons  can  be  seated  at  a  dinner  table.  Ans.  720. 

181.  The  Number  of  Combinations.  —  To  find  the 
number  of  combinations  of  n  different  tilings  taken  r  at  a 
time. 

The  number  of  permutations  of  n  things  taken  r  at  a  time 
is 

n{n  -  l)(n  -  2) (?i  —  r  +  1).    (Art.  180). 

But  each  combination  of  r  things  taken  r  at  a  time  will 
make  I  r  permutations,  by  (2)  of  Art.  180  ;  therefore  there 
are  |  r  times  as  many  permutations  as  combinations.    Hence, 
calling  „Cr  the  required  number  of  combinations,  we  have 
n         W(w-  1)(W_  2) {n-r  +  1) 

This  formula  for  nCr  may  also  be  written  in  a  different 
form  ;  for  if  we  multiply  the  numerator  and  the  denominator 
by  the  product  of  the  natural  numbers  from  1  up  to  n  —  r, 
it  becomes 

c   _  n(n-l)(n-2) (n-r+1)  (n-r) 2-1 

[r-  (n-r)(n-r-l) 2-1 

The  numerator  is  now  the  product  of  the  natural  numbers 

from  n  to  1,  or  is  In  (Art.  180);    the  denominator  is  the 

product  of  the  natural  numbers  from  /•  to  1,  and  from  n  —  r 

to  1.     Hence  we  have  . 

\n 

•C-^h-r-     •     •'     •     '     -(2) 

It  will  be  convenient  to  use  (1)  for  „Cr  in  all  cases  where 
a  numerical  result  is  required,  and  (2)  when  it  is  sufficient 
to  leave  it  in  an  Algebraic  shape. 

Note  1.  —  If  in  (2)  we  put  r  =  n,  we  have 

C  I—    =    1  • 

"  "       |n  [0       |0' 

but  „Cn  =  1,  so  that  if  the  Formula  is  to  be  true  for  r  —  »,  the 
symbol  [0  must  be  considered  as  equivalent  io  1. 


THE  NUMBER   OF  COMBINATIONS.  389 

The  number  of  combinations  of  n  things  taken  r  at  a  time 
is  the  same  as  the  number  of  them  taken  n  —  r  at  a  time. 
For  the  number  taken  n  —  r  ut  a  time  is,  from  (2), 

\n  \n 

„C»-r  =  j rJ-7 r  =  |       L    |    ,      •    (3) 

\n  —  r  \n  —  (n  —  r)        w  —  r  \r 

which  =  „Cy,.,  from  (2). 

The  truth  of  this  proposition  is  also  evident  from  the 
consideration  that  for  every  different  group  of  r  things  taken 
out  of  n  things  there  is  always  left  a  different  group  of 
n  —  r  things.  Hence  the  number  of  groups  of  r  things  out 
of  n  must  be  the  same  as  the  number  of  groups  of  n  —  r 
things.     Such  combinations  are  called  complementary. 

Note  2.  —  Put  r  =  n;  then  from  (2)  and  (3),  »C„  =  „C0  =  1. 

The  proposition  just  proved  is  useful  in  enabling  us  to 
abridge  Arithmetic  work.     Thus, 

1.  Required  the  number  of  combinations  of  20  things 
taken  18  together. 

The  required  number  is  the  same  as  the  number  taken  2 
together.  90  v  1<> 

''•    »cs  =  T#T  =  190- 

If  we  had  used  the  formula  o0Ci8i  we  should  have  had  to 
reduce  an  expression  whose  numerator  and  denominator 
each  contained  18  factors. 

2.  From  12  books,  in  how  many  ways  can  a  selection  of  5 
be  made  when  one  specified  book  is  always  excluded? 

Since  the  specified  book  is  always  to  be  excluded,  we  have 
to  select  the  5  books  out  of  the  remaining  11.     Hence 

=  11.10. 9-8-7  =  462< 
11   5  1-2-3-4.5 

3.  How  many  combinations  may  be  made  of  10  letters 
taken  6  at  a  time?  Ans.  210. 

4.  From  11  books,  in  how  many  ways  can  a  selection  of 
4  be  made?  Ans.  330. 


390      TO  DIVIDE  m  +  n   THINGS   INTO   TWO  CLASSES. 

182.  To  Divide  m  +  n  Things  into  Two  Classes. 
—  To  find  the  number  of  xcays  in  which  m  -f-  n  different 
things  can  be  divided  into  two  classes,  so  that  one  may  contain 
m  and  the  other  n  things. 

This  is  equivalent  to  finding  the  number  of  combinations 
of  m  +  n  things  m  at  a  time,  for  every  time  we  select  one 
group  of  m  things,  we  leave  a  group  of  n  things  behind. 
Hence  by  (2)  of  Art.  181, 

Ihe  required  number  =  . 

[m  \n 

In  a  similar  manner  it  may  be  shown  that  the  number  of 
ways  in  which  m  +  n  +  p  different  things  can  be  divided 
into  three  classes  containing  m,  n,  p  things  respectively  is 

\m  +  n  +  p 

[m  In  \p 

1.  There  are  three  bookshelves  capable  of  containing  14, 
22,  and  24  books ;  in  how  many  ways  can  60  books  be 
allotted  to  the  shelves? 

Here  we  have  to  divide  GO  things  into  groups  of  14,  22, 
and  24  things. 

160 

Hence  the  required  number  =  == . 

1  [14  [22  1 24 

2.  From  7  Englishmen  and  4  Americans  a  committee  of 
6  is  to  be  formed,  containing  2  Americans  ;  in  how  many 
ways  can  this  be  done? 

Here  we  have  to  choose  2  Americans  out  of  4,  and  4 
Englishmen  out  of  7.  The  number  of  ways  in  which  the 
Americans  can  be  chosen  is  4C2 ;  and  the  number  of  ways  in 
which  the  Englishmen  can  be  chosen  is  7C4.  Each  of  the 
first  groups  can  be  associated  with  each  of  the  second 
Hence  the  required  number  of  ways  is 

(C.XiCl  =  lXI  =  J!      =210. 

EB     l*E     BB15 


PERMUT. 


■ITIONS  OF  n  THINGS  NOT  ALL  DIFFERENT.   39] 


3.    In   how  many  ways  can  the   52   cards  in  a  pack  be 

^2 
divided  among  4  players,  each  to  have  13?  Ans.  . 

183.  Permutations  of  n  Things  not  all  Different. 

—  To  find  the  number  of  permutations  of  n  things  taken  all 
at  a  time,  when  they  are  not  all  different. 

Let  there  be  n  letters ;  and  suppose  p  of  them  to  be  a,  q 
of  them  to  be  b,  r  of  them  to  be  c,  and  the  rest  to  be 
unlike. 

Let  P  be  the  required  number  of  permutations.  If  in 
any  one  of  the  actual  permutations,  the  p  letters  a  were  all 
changed  into  p  letters  different  from  each  other  and  from 
all  the  rest,  then  from  this  single  permutation,  without  alter- 
ing the  position  of  any  of  the  remaining  letters,  we  could 
form  \p  new  permutations.  Hence  if  this  change  were  made 
in  each  of  the  P  permutations,  there  would  be  P  X  \p  per- 
mutations. 

Similarly,  if  in  any  one  of  these  new  permutations,  the  q 
letters  b  were  changed  into  q  letters  different  from  each  other 
and  from  all  the  rest,  then  from  this  single  permutation  we 
could  form  \q  new  permutations.  Hence  the  whole  number 
of  permutations  would  now  be  P  X  \p  X  \q. 

In  like  manner,  if  the  r  letters  c  were  also  changed  so  that 
no  two  were  alike,  the  total  number  of  permutations  would 
be  P  X  \p  X  \q  X  |r.  But  this  number  must  be  equal  to 
the  number  of  permutations  of  n  different  things  taken  all 
together,  which  is  \n.     Hence 

P  X  [p  X   \q  X   [r  =  \n. 


And  similarly  any  other  case  may  be  treated. 

1.    How  many  different  permutations  can  be  formed  out  of 
the  letters  of  the  word  Mississippi  taken  all  together? 


392  EXAMPLES. 

Here  we  have  11  letters,  of  which  4  are  i,  4  are  s,  and  2 

are  p. 

|H 
.-.     P=  H— -  =  ILIO.9.7.5  =  34650. 

mm 

2.  TTow  many  different  permutations  can  be  made  out  of 
the  letters  of  the  word  assassination  taken  all  together? 

Arts.  10810800. 

EXAMPLES. 

Reduce  each  of  the  following  fractions  to  a  continued 
fraction,  and  find  the  fourth  convergent  to  each. 

.832  .-,11111       157 

1.    .  Ans.  5  H —  — ;    . 

159  4+  3+  2+  1+3       30 

2  729  _L_LJ__L  J_  _L_L  L  11 

2318'  3+5+1+1+3+2+1+5'    35* 

3  3029  J_J_J_J_J_J_J_J_.     ™. 

3+  3+ 3+  G+  1+  2+  1+  10'    208' 

4.  A  metre  is  39.37079  inches  ;  show  by  the  theory  of 
continued  fractions  that  32  metres  is  nearly  equal  to  35 
yards. 

5.  A  kilometre  is  very  nearly  equal  to  .62138  miles  ;  show 
by  the  theory  that  the  fractions  §,  |f,  §f,  i';,i.!  are  successive 
approximations  to  the  ratio  of  a  kilometre  to  a  mile. 

G.  Two  bells  begin  to  ring  together;  the  one  rings  12 
times  in  7  minutes,  and  the  other  17  times  in  9  minutes; 
what  strokes  most  nearly  coincide  in  the  first  half-hour? 

Ans.  The  49th  of  the  first  with  the  54th  of  the  second. 

7.  Express  27.321061  days,  the  average  period  of  the 
moon's  revolution,  by  the  nearest  equivalent  fraction  whose 
denominator  is  less  than  1000.  Ans.  27$$f  days. 

8.  Find  a  series  of  fractions  converging  to  .2422642,  the 
excess  in  days  of  the  true  tropical  year  over  365  days. 

AriQ    X      1      JL      39        47        321       r>+« 


EXAMPLES.  393 

9.  The  relative  sidereal  periods  of  Venus  and  the  earth 
are  224700  days  and  3G525G  days  :  find  a  series  of  fractions 
converging  to  fffHt-  Ans'  *>  h  h  h  fs,  Iff'  etc- 

The  5th  convergent,  which  precedes  the  large  quotient  29,  is  a  very 
near  approximation,  the  error  being  less  than  [Art.    17(5,    (3)]. 

In  consequence  of  this,  a  transit  occurs  after  a  period  of  S  years,  and 
then  again  not  till  after  235  years  have  been  completed. 

10.  Express  \/l9  as  a  continued  fraction,  and  find  a  series 
of  fractions  approximating  to  its  value,  and  show  that  the 
7th  convergent  will  give  the  true  value  to  at  least  4  places  of 
decimals. 

Ans.  ^19  =  4+    — t^-t^-^-^-TT- >   con" 

v  ^2+1+  3+  1+  2+  8  + 

vergents  are  f,  f ,  V.  **,  ft.  W,  W»  •  •  j_-  •  • 

11.  Show  that  the  9th  convergent  to  \J33  will  give  the 
true  value  to  at  least  6  places  of  decimals. 

12.  Find  the  limits  of  the  error  when  %y   is  taken  for 

^23-  Ans. and  , ?— -. 

44  x  49  44  x  93 

13.  Find  the  value  of  - — ■  - —  - — 

1+  2+  1+  2  + 

Ans.  Positive  root  of  ar  +  2x  =  2. 

14.  Find  the  value  of  ±.  ±.  -L  _L  J-  _L 

Ans.  Positive  root  of  Ix1  +  8x  =  3. 

15.  How  many  different  numbers  can  be  formed  by  using 
six  out  of  the  nine  digits,  1,  2,  3, 9?       Ans.  G0480. 

16.  Required  the  number  of  changes  which  can  be  rung 
upon  12  bells  taken  all  together.  Ans.  479001G00. 

17.  Required  the  number  of  combinations  of  24  different 
letters  taken  4  at  a  time.  Ans.  10G26. 

18.  Out  of  14  men,  in  how  many  ways  can  11  be  chosen? 

Ans.  364. 


394  EXAMPLES. 

19.  How  man}'  different  products  can  be  formed  with  any 
three  of  the  figures  1,  3,  5,  7,  9?  Ans.  10. 

20.  In  how  many  ways  can  6  copies  of  Horace,  4  of 
Virgil,  and  3  of  Homer  be  given  to  13  boys,  so  that  each 
boy  may  receive  a  book?  Ans.  G0060. 

21.  Out  of  7  consonants  and  4  vowels,  how  many  words 
can  be  made  each  containing  3  consonants  and  2  vowels? 

Ans.  25200. 

22.  How  many  parties  of    12  men  each  can  be  formed 

from  a  company  of  GO  men?  ]60 

Ans. 


II?  II? 

23.  Out  of  12  Republicans  and  1G  Democrats,  how  many 
different  committees  could  be  formed,  each  consisting  of  3 
Republicans  and  4  Democrats?  *         llj  H^ 

'  |3  [9       |4  J12/ 

24.  Out  of  10  consonants  and  4  vowels,  how  many  words 
can  be  formed,  each  containing  3  consonants  and  2  vowels? 

Ans.  86400. 

25.  There  are  10  candidates  for  G  vacancies  in  a  commit- 
tee :  in  how  many  ways  can  a  person  vote  for  6  of  the  can- 
didates? Ans.  210. 

26.  In  how  many  ways  can  a  cricket  eleven  be  chosen  out 
of  fourteen  players?  Ans.  364. 

27.  In  how  many  ways  could  2  ladies  and  2  gentlemen  be 
chosen  to  make  a  set  at  tennis  from  a  party  of  4  ladies  and 
G  gentlemen?  Ans.  90. 

28.  In  how  many  ways  could  2  ladies  and  2  gentlemen  be 
chosen  to  make  a  set  at  tennis  from  a  party  of  G  ladies  and 
8  gentlemen  ?  Ans.  420. 

29.  From  6  ladies  and  5  gentlemen,  in  how  many  ways 
could  you  arrange  sides  for  a  game  of  croquet,  so  that  there 
would  be  2  ladies  and  one  gentleman  on  each  side' 

^'•s'-(#7ror1800- 

30.  Out  of  6  ladies  and  <s  gentlemen,  how  man?  different 


EXAMPLES. 

parties 

can 

be 

formed, 

each  consisting 

of  3 

ladies   i 

gentlemen  ? 

Ans, 

^x 

•|8|8X 

395 

,nd  4 
JL 

31.  How  many  different  permutations  can  be  made  out  of 
the  letters  of  the  word  Heliojjolis?  Ans.  453600. 

32.  How  many  out  of  the  letters  of  the  words,  Ecclesias- 
tical, Papatoitoi?  .  „  110 

1  Ans.  454053600  ;  -!=-. 

(|2)5 

33.  If  the  number  of  permutations  of  n  things  taken  4 
together  is  equal  to  12  times  the  number  of  permutations  of 
n  things  taken  2  together  ;  find  n.  Ans.  6. 

34.  In  how  many  ways  can  a  party  of  6  take  their  places 
at  a  round  table  ?  Ans.  60. 

35.  How  many  words  of  two  consonants  and  one  vowel 
can  be  formed  from  6  consonants  and  3  vowels,  the  vowel 
beiug  the  middle  letter  of  each  word  ?  Ans.  90. 

36.  How  many  words  of  6  letters  may  be  formed  with 
3  rowels  and  3  consonants,  the  vowels  always  having  the 
even  places?  Ans.  36. 

37.  There  are  15  boat-clubs ;  2  of  the  clubs  have  each  3 

boats  on  the  river,  5  others  have  2,  and  the  remaining  8  have 

one  :  find  the  number  of  ways  iu  which  a  list  can  be  formed 

of   the  order   of   the  24  boats,  observing  that  the  second 

boat  of  a  club  cannot  be  above  the  first.  |24 

Ans. 


38.  How  many  numbers  can  be  formed  with  the  digits  1, 
2,  3,  4,  3,  2,  1,  so  that  the  odd  digits  always  occupy  the  odd 
place?  Ans.  18. 

39.  In  how  many  ways  can  5  prizes  be  given  away  to  4 
boys,  when  each  boy  is  eligible  for  all  the  prizes? 

Ans.  1024. 

40.  A  man  has  6  friends  :  in  how  many  ways  may  he 
invite  one  or  more  of  them  to  dinner?  Ans.  63. 


396     DEFINITIONS.  —  INDETERMINATE   COEFFICIENTS. 


CHAPTER     XIX. 

.INDETERMINATE*    COEFFICIENTS  — PARTIAL 

FRACTIONS  —  BINOMIAL    THEOREM. 

184.  Definitions.  —  An  Algebraic  expression  is  said  to 
be  developed,  when  it  is  transformed  into  a  series  the  sum  of 
all  the  terms  of  which  is  equal  to  the  given  expression.  The 
series  itself  is  called  the  development  of  that  expression. 

When  the  sum  approaches  a  limit  as  the  number  of  terms 
is  increased,  the  series  is  said  to  be  convergent.  When  the 
sum  does  not  approach  a  limit  as  the  number  of  terms  is 
increased,  the  series  is  said  to  be  divergent. 

Thus,  the  series  in  Art.  165  are  convergent  series.  The 
series  1  +  2  +  3  +  etc.  to  oo  is  divergent  because  there  is 
no  limit  to  the  sum  of  its  terms. 

Any  irreducible  fraction  may  be  developed  into  an  infinite 
series  by  dividing  the  numerator  by  the  denominator.  Thus, 
by  division  we  find  that  the  fraction 

1  - 


1  —  x  +  x' 


=  1  —  x*  —  x*  +  xb  +  xb  —  x*  —  etc. 


The  method  which  is  the  most  frequently  used  for  devel- 
oping expressions,  is  that  of  Indeterminate  Coefficients.  It 
consists  in  assuming  for  the  required  development  a  series 
with  unknown  coefficients,  and  then  finding  the  values  of 
these  coefficients. 

185.  Indeterminate  Coefficients.  —  (1)  If  the  infinite 

series  a0  +  a}x  +  atf?  +  aAx8  + is  equal  to  zero 

for  every  finite  value  of  x  for  which  the  series  is  convergent 
then  each  coefficient  must  be  equal  to  zero. 

*  Culled  alao  «><</.  u  rminati  and  undetermined  coefficients. 


DEVELOPMENT  BY  INDETERMINATE  COEFFICIENTS.    397 

For  since  the  equation, 

a0  +  axx  +  a2x*  +  aaxs  +  etc.  =  0, 
is  true  for  fill  values  of  x  which  make  it  convergent,  it  must 
be  true  when  x  =  0.     Putting  x  =  0,  it  becomes  a0  =  0. 
Removing  o0,  the  equation  becomes 

a±x  +  a2x2  +  a3^3  +  etc  =  0. 
Dividing  by  a;, 

«i  +  a2a;  +  a3a2  +  etc.  =  0. 
This  equation  is  true  for  all  values  of  x  which  make  it 
convergent ;    hence  it  must  be  true  when  x  —  0.      Putting 
x  =  0,  it  becomes  a2  =  0. 

Continuing  the   process,  we  find   a.2  =  0,   a8  =  0,   etc. 
indefinitely. 

(2)   If  tivo  infinite  series  are  convergent  and  equal  to  each 
other  for  every  finite  value  of  the  variable,  the  coefficients  of 
the  like  powers  of  the  variable  in  the  tivo  series  are  equal. 
Let  the  two  equal  series  be 

a0  +  «!#  +  a2x2  +  asx3  +  etc., 
b0  +  bjX  +  b2x2  +  \xz  +  etc. 
Subtract  the  second  from  the  first ;  then 

«0  -  h  +  (a1  -  bx)x  +  (o2  —  b2)x2  +  etc.  =  0. 
Since  this  equation  is  true  for  all  values  of  x,  the  coeffi- 
cients of  the  different  powers  of  x  must  all  be  zero  by  (1) ; 
hence 

a0  —  b0  =  0,     ax  —  bx  —  0,     a2  —  b2  =  0,     etc.  ; 
that  is,  a0  =  b0,     a,  =  b^     a2  =  b2,     etc. 

The  application  of  these  equations  is  known  under  the 
name  of  equating  coefficients  of  like  poivers  of  x. 

186.  Development  by  Indeterminate  Coefficients. 

—  The  theorem  given  in  Art.  185  is  sometimes  referred  to 
as  the  Principle  *  of  Indeterminate  Coefficients.     It  is  applied 

*  Called  also  method  of  indeterminate  coefficients. 


398    DEVELOPMENT  BY  INDETERMINATE  COEFFICIENTS. 

to  the  development  of  expressions  in  powers  of  the  variable. 
The  application  of  the  principle  is  best  .seen  in  the  following 

examples.* 

2  4-  x2 

1.    Expand  — into  a  series  of  ascending  powers 

1  +  x  —  x2 

of  x  as  far  as  the  term  involving  a;6. 

It  is  plain  that  the  development  is  possible,  for  we  have 

seen  (Art.  184)  that  any  fraction  may  be  expanded  into  a 

series  by  dividing  the  numerator  by  the  denominator.     Let 

us  then  assume 

2  +  ^     =  A  +  Bx  +  Cx2  +  Dx*  +  Ex*  +  Fx5  +  etc.,   (1 ) 
1  +  X  —  x1 

where  A,  B,  C,  etc.  are  independent  of  x,  but  depend  upon 
the  constants  of  the  fraction,  and  its  form.  It  is  now 
required  to  find  such  values  for  the  constants,  A,  B,  C,  etc. 
as  will  make  the  assumed  development  true  for  all  values 
of  x. 

Clearing  (1)  of   fractions,  and  collecting  the  terms  with 
the  same  powers  of  a;,  we  have 

2+x2  =  A  +  {A  +  B)x  +  {-A  +  B+C)x2+(-B+C+D)x* 
+  (-C+D  +  E)xi  +  (-J5  +  E  +  F)x5  +  etc. 

Equating  the  coefficients  of  the  like  powers  of  x  (Art.  185), 
we  have 

A  =  2. 

A  +  B  =  0,  which  gives  B  =  -A  =  -2. 
-A  +  75  +  C  =  1,  which  gives  C  =  1   +  A  -  B  =  5. 
_73  +  c  +  T)  =  0,  which  gives  D  =  B  -  G  =  -   7. 
_C  +  Z>  +  E  =  0,  which  gives  E  =  (7  -  D  =      12. 
-7J  +  #  +  F  =  0,  which  gives  F  =  2>  -  E  -  -19. 

*  In  these  applications  the  Beriea  are  alwaya  regarded  as  convergent 


DEVELOPMENT  BY  INDETERMINATE  COEFFICIENTS.    399 

Substituting   these  values   of   the   coefficients   in    (1),  it 
becomes 

—  =  2  -  2a;  +  5a2  -  7x3  +  12a4  -  19a5  +  etc. 


1  +  x  -  x1 

In  applying  the  method  of  indeterminate  coefficients,  the 
form  of  the  given  expression  will  determine  what  powers  of 
x  must  be  assumed.  In  the  example  just  solved,  we  assumed 
the  series  to  be  arranged  according  to  the  ascending  powers 
of  x  beginning  with  x° ;  but  in  many  series  this  is  not  the 
case,  and  the  error  in  the  assumed  series  will  then  appear, 
either  by  an  absurd  result,  or  by  the  coefficients  of  those 
terms  which  do  not  exist  in  the  actual  series  reducing  to 
zero.  By  dividing  the  term  of  the  numerator  having  the 
lowest  power  of  x  by  the  term  of  the  denominator  having 
the  lowest  power  of  a;,  we  may  always  determine  the  expo- 
nent of  x  in  the  first  term  of  the  series.  We  may  then 
assume  a  series  beginning  with  this  exponent  of  x,  and 
arranged  according  to  the  ascending  powers  of  x  as  usual. 
Thus, 

2.    Expand  into  a  series. 

1         3x  -  x2 

If  we  should  assume  the  series  to  be  A  +  Bx-\-  Ox2  +  etc., 
on  clearing  of  fractions,  and  proceeding  to  equate  the  coeffi- 
cients of  like  powers  of  a?,  we  would  have  no  term  on  the 
right  corresponding  to  1  on  the  left ;  so  that  we  would  have 
1  =  0,  an  absurd  result.  Hence  we  infer  that  the  expres- 
sion cannot  be  developed  according  to  the  ascending  powers 

of  x  beginning  with  x°. 

r-i 
But,  dividing  1  by  3x,  we  get  ■ —  for  the  first  term  of  the 
o 

quotient ;  hence  we  may  assume  a  series  beginning  with  a>-1, 

and  the  exponents  increasing  by  unity.     Thus, 

■ — =  Ax'1  +  B  +  Cx  +  Dx1  +  Ex3  +  etc. 

3x  —  x'2 


400  DEVELOPMENT  BY  INDETERMINATE  COEFFICIENTS. 

Clearing  this  of  f motions,  and  collecting  the  terms  with 
the  same  powers  of  x,  we  have 

1  =  BA  +  (BB  -  A)x  +  (3(7  -  B)x2  +  (3D  -  C)x3  +  etc. 

Equating  the  coefficients  of  the  like  powers  of  x,  we  find 

A  =  |,     5  =  i,     C  =  aV,     ^  =  JT,     etc. 

Therefore =  —  +  -  +  —  +  —  +  etc. 

3a;  -  x2         3    T9T27T81T 


3.    Expand  \l+x  into  s 


a  series. 


Assume  \l+x=A  +  Bx+  Cx2  +  Da8  +  etc. 
Squaring, 

l+a;=  A2+  2ABx  +  (B2+  2AC)x*+  (2BC  +  2AD)x*+  etc. 
Equating  the  coefficients  of  like  powers  of  x, 
A2  =  1,         .-.     A  =      1. 
2AB  =  1,         .-.     5  =      f 
£2  +  2^1(7  =  0,         .-.     C  =  -J. 
25(7  +  2.4Z)  =  0,         .-.     D  =    TV,     and  so  on. 


Note.  —  Since  A2  =  1,  A  =  ±1 ;  we  might  take  the  value  .4  =  —  1 . 
in  which  case  we  would  find  B  —  —\,  C  =  $,  D  — — rb-,  and  so  on, 

so  that  \J\  +  x  =  _l_?  +  ?2_£3_  etc.     (See  Note  on  Art.  112.) 

2  T  S       16 

Expand  the  following  to  four  terms : 

4.  L±_??  ^ns.  j  +  rjr  +  15aji  +  45rs# 
1  —  3a* 

5.  _J  +  2x  1  +  3.*:  +     4.r  +     7.v8. 


6.  vT^.  1  -  •''  -  £  - 

2      a      16 


PARTIAL   FRACTIONS.'  401 


PARTIAL    FRACTIONS. 

187.  Partial  Fractions.  Case  I.  —  When  the  factors  of 
the  denominator  are  all  unequal.  When  the  denominator  of  a 
fraction  can  be  resolved  into  factors,  we  may  by  the  principle 
of  indeterminate  coefficients  decompose  the  fraction  into  the 
sum  of  two  or  more  simpler  fractions,  called  partial  frac- 
tions, the  denominators  of  which  are  the  factors  of  the  given 
denominator.  We  shall  give  a  few  examples  illustrating  the 
decomposition  of  a  rational  fraction  into  partial  fractions. 
For  a  general  discussion  of  the  subject,  the  student  is 
referred  to  Treatises  on  the  Theory  of  Equations,  and  on 
the  Integral  Calculus.* 

Kr    11 

Resolve  — — into  partial  fractions. 

2x2  +  x  —  6 

Since  the  denominator  2x'2  -f  x  —  6  =  («  +  2)  (2a  —  3), 
we  assume        5a;  _  n  A  B  (1) 

2x2  +  a;-6       x  +  2       2x  -  3' 
where   A   and   B   are    independent    of    x,    and   at   present 
unknown.     We  have  then  to  find  such  values  for  A  and  B 
as  will  make  (1)  an  identity,  that  is,  true  whatever  may  be 
the  value  of  x.     Clearing  (1)  of  fractions,  it  becomes 

5x  -  11  =  A(2x  -  3)  +  B(x  +  2)      .     .    (2) 
Since  (2)  is  identically  true,  we  may  equate  the  coefficients 
of  like  powers  of  x  ;  thus, 

2A  +  B  =  5,     -SA  +  2B  =  -11  ; 
whence         .4  =  3,     and     B  =  —  1,         which  in  (1)  gives 

5x  -  1 1      _       3  1 

2jt  +  x  —  6       x  +  2       2aj  —  3* 

Second  Method.  In  practice,  in  this  case,  there  is  a  simpler 
method  of  finding  the  values  of  A,  B,  etc.  than  by  equating 

*  See  Todhunter's  Theory  of  Equations,  Chap.  XXIV.,  Integral  Calculus, 
Chap.  II.,  also  Serret's  Cours  d'Algebre  Superieure. 


402  CASE  II. 

the  coefficients.     Since  A  and  B  are  independent  of  x,  we 

may  give   to  x  any   value   we    please.      If   in    (2)   we   put 

x  4-  2  =  0,  or  a:  =  —2,  then  we  have  immediately 

-10  -11  =  -1A,     or     ^1  =  3. 

In  the  same  way,  putting  2.»;  —  3  =  0,  or  X  =  |,  we  get 

¥  -  11  =  (I  +  2)-B,     or     B  =  -1,     as  before. 

188.   Case  II.  —  When  the  factors  of  the  denominator  are 

equal. 

2  3^2 

1.    Resolve — -  into  partial  fractions. 

[x  +  2)3 

Assume   2  ~  3x*  =  -A-  +         *         +         O  () 

(a  +  2)3       x  +  2        (as  +  2)*  T  (a  +  2)3  V 

Rem.  —  All  the  factors  of  the  given  denominator  must  be  taken 
as  denominators  of  the  assumed  fractions.  Thus,  in  this  example, 
the  denominators  of  the  partial  fractions  may  he  (,c  +  2),  (x  +  2)2, 
(x  +  2)s.  It  is  therefore  necessary  to  assume  fractions  with  each  of 
these  denominators,  so  as  to  include  every  possible  case  which  can 
produce  the  given  fraction,  although  A  or  B  may  reduce  to  zero. 

Clearing  (1)  of  fractions,  it  becomes 

2  -  3z2  =  A(x  +  2)2  +  B(x  +  2)  +  C. 
Equating  the  coefficients  of  like  powers  of  x, 
±A  +  2B  +  C  =  2, 

4A  +  B  =  0,     and     .1  =  —3. 
Solving  these  three  equations,  we  get 

A  =  -3,     B  =  12,     C  =  -10, 
which  in  (1)  gives 

2  -  3.-C2  _  3  12  10 


(sb  +  2)3  x  +  2        (.»•  +  2)a        (a:  +  2)3' 

NOTE.  —  In  this  case  we  cannot  iind  the  values  of  .1,  />',  C,  etc.  by 
the  second  method  of  Case  I.,  but  have  to  employ  the  first.  When 
both  equal  and  unequal  factors,  however,  occur  in  the  denominator, 

both  methods  may  be  combined  to  advantage,  as  in  the  following 
example. 


CASE  III.  403 


2.    Resolve  — — into  partial  fractions. 

x'\l  +  x)* 

Assume 

A  +  ?  +  _^_  + *>        (See  Rem.)     (1) 


x\l  +  X)2       X       x2       1+x       (1  +  x)2 

Clearing  (1)  of  fractions,  it  becomes 
Bx-l  =  Ax(l+xy  +  B(l+x)2  +  Cx2(l+x)  +  Dx*   .   (2) 

Here  we  may  use  the  second  method  of  Case  I.  as  follows  : 

Putting  x  =      0,  we  find  B  =  —  1. 

Putting  x  =  —  1,  we  find/)  =  —4. 

To  find  A  and  C,  equate  the  coefficients  of  x2  and  a3; 
thus,  equating  coefficients  of  x2, 

0  =  -2  A  +  £  +  0  +  #, 

or  2^1  +  C  =  5 (3) 

Equating  coefficients  of  a:8, 

yl  +  (7  =  0 (4) 

Solving  (3)  and  (4),  we  find 

A  =  5,     C  =  -5, 
which  in  (1)  gives 

3x  -  1      =  5  _  J_  _       5       _  4 

ar(l  +  x)2       x       x2        1  +  x        (1  +  a-)'2' 

189.     Case    III.  —  When   some   of  .  the   factors    of   the 

denominator  are  of  the  second  degree. 

42  19a; 

1.    Resolve into  partial  fractions. 

(oF  +  l)<a,_4) 

Here  we  have  the  factor  x2  +  1  for  the  denominator  of 

one  of  the  partial  fractions  ;  the  numerator,  therefore,  may 

contain  the  first  power  of   .r,  as  well  as  the   zero   power  ; 

therefore  assume 

42  -  19.«         _  Ax  +  B  C 


(x2  +  l)(a--  4)         x*+l         b-4 

.-.     42  -  19a  =  (Ax  +  B)(x  -  4)  +  C\x2  +  1). 


404  CASK  TIL 

Putting  x  =  4,  we  find  G  =  —  2. 

Equating  coefficients  of  a;3,      0=J.+  (7;  .•.    A=2. 

Equating  the  absolute  terms,  42=—  4B+  C ;      .-.    B=  — 11. 

42  -  19a;  2a;  -  11  _        2 

(a;2  +  1)  (x  -  4)  ~~    x1  +  1         a  -  4' 

In  each  of  these  four  examples  the  numerator  has  been  of 
a  lower  degree  than  the  denominator.  When  this  is  not  the 
case  we  may  by  division  (Art.  81)  reduce  the  fraction  to 
the  sum  of  an  integral  expression  and  a  fraction  whose 
numerator  is  of  a  lower  degree  than  the  denominator. 

6#3  -f-  5a;2  7 

2.    Resolve    '  into  partial  fractions. 

3a;2  -  2x    -  1  ' 

Bydivision,  ^  +  5x*  ~  7  =  2x  +  3  +        8*  ~  4       . 
J  3a;2  -  2x    -  1  3x-  -  2x  -  1 

Resolving  this  fraction  into  partial  fractious  as  in  (1), 
Art.  187, 

8a;  -  4  5  1 

3a;2  —  2x  —  I  3a;  -f  1       x  —  l" 

•       **  +  ^    -  -'   =  2*  +  3  +  _«_  +       > 


3a;2  -  2a:    -  1  3a;  +  1        x  —  1 

Resolve  into  partial  fractions  : 

Q     8a;  -  4  .5,3 

3.    —„ .  Ans. 


x2  —  4  x  +  2       x  -  2 

23a;  -  11a;2  1  1  1 


(2x  -  1)(9  -  a;2)  2a;  -  1        3  +  *       3  -  a; 

33,2  +  x  _  2  1  5  I 

(«_  2)«(l  -2a;)"  3(1 -2a;)       3(a--2)       (a--2)2' 

2G.r2  +  208.C 
(a;2+  !)(»  +  5)' 


\x  +  5         .r  +   1   / 


BINOMIAL    THEOREM.  405 


INOM1AL    THEOREM. 


190.  Positive  Integral  Exponent.  —  The  method  of 
raising  a  binomial  to  any  power  by  repeated  multiplication 
has  been  explained  in  Art.  109.  We  shall  now  prove  a 
formula  known  as  the  Binomial  Theorem,*  by  which  any 
binomial  can  be  raised  to  any  power  without  the  labor  of 
actual  multiplication. 

To  prove  the  Binomial  Theorem  for  a  positive  integral 
exponent. 

By  actual  multiplication  we  obtain 
(x  +  a)  (x  +  b)  =  x2  +  (a  +  b)  x  +  ab, 
(x+a)  (x+b)  (x+  c)  —  x3+  (a+b  +  c)x2  +  (ab  +  ac  +  bc)x  +  abc. 

In  these  results  we  see  that  the  following  laws  hold  : 

1.  The  number  of  terms  on  the  right  side  is  one  more  than 
the  number  of  the  binomial  factors  on  tJie  left  side. 

2.  The  exponent  of  x  in  the  first  term  is  the  same  as  the 
number  of  binomial  factors,  and  decreases  by  one  in  each 
successive  term. 

3.  The  coefficient  of  the  first  term  is  unity ;  of  the  second 
term,  the  sum  of  the  letters  a,  b,  c;  of  the  third  term,  the  sum 
of  the  products  of  the  letters  a,  b,  c,  taken  tioo  at  a  time;  and 
the  fourth  term  is  the  product  of  all  the  letters. 

We  shall  now  prove  by  induction  (Art.  170)  that  these 
laws  always  hold  whatever  be  the  number  of  binomial  factors. 

Suppose  these  laws  to  hold  for  n  —  1  binomial  factors,  so 
that 

{x+a)  (x+b)  .  .  (x+Jc)  =xn-1+Axn-'2+Bxn-3+Cxn-i+  .  .  K,  (1) 
where  A  =  a-\-b-\-c-\-.  ...+&,  the  sum  of  the  second 
terms, 

B  =  ab  -\-  ac  +  be  + ,  the  sum  of  the  products 

of  these  terras  taken  two  at  a  time. 

C  =  abc  +  abd  +■ ,  the  sum  of  the  products  of 

these  terms  taken  three  at  a  time. 

K  =  abed k,  the  product  of  all  these  terms. 

*  This  theorem  was  discovered  by  Newtou. 


406  POSITIVE   INTEGRAL   EXPONENT. 

Multiply  both  sides  of  (1)  by  another  factor  (x  +  J)  ;  thus, 
(x+a) (x+b)  ....  (x+k)(x+-l)=x"+(A  +  l)xa-1 

+  (B+Jl)x*-S+(C+Bl)ar*+ +KI .  .  •  (2) 

Now  A  +  l=a  +  b  +  c  + -f  k  +  I 

=  the  sum  of  all  the  terras  a,  6,  c, ?• 

5  +  Al  =  a&+ac+6c+  .  .  .  +al+bl+  cl  +  .  .  .  +  W. 
=  the  sum  of  the  products  taken  two  at  a  time. 

C  +  Bl  =  abc  +  dbd  +  .  .  .  +  aW  +  ad  +  6d  +  .  .  . 
=  the  sum  of  the  products  taken  three  at  a  time. 


Kl  =  <(bcd  .  .  .  kl  =  the  product  of  all  the  terms 
o,  &,  c,  .  .  .  .  I. 

Also  the  exponent  of  x  in  the  first  term  is  the  same  as  the 
number  of  binomial  factors,  and  decreases  by  1  in  each  suc- 
cessive term. 

Hence  if  the  laws  hold  when  n  —  1  factors  are  multiplied 
together,  they  hold  when  n  factors  are  multiplied  together ; 
but  they  have  been  proved  to  hold  for  3  factors,  therefore 
they  hold  for  4  factors,  and  therefore  for  5  factors,  and  so 
on,  generally,  for  any  number  whatever. 

Now  let  6,  c,  d, Z,  each  =  a  ;  then  the  binomial 

factors  are  all  equal,  and  the  first  member  of  (2)  becomes 

(x+a)  (x+a)  .  .  .  =  (x+a)  taken  n  times  as  a  factor  =  (x+a)n ; 

and  the  second  member  becomes 

A  +  I  =  a  +  a  +  a  +-....  =  a  taken  n  times  =  na. 

B  +  Al  =  an  +  aa  +  .  .  .  =  a'2  taken  as  many  times  as  there 

are  combinations  of  n  letters  taken  2  at  a  time  =  n^n  ~ — '- 

I2 
(Art.  181). 

C  +  Bl  =  aaa  +  ami  + =  a?  taken  as  many 

times  as  there  are  combinations  of  n  letters  taken  ;i  at  a 
time  =  n(n-l)(n-2) 

Li 


POSITIVE   INTEGRAL   EXPONENT.  407 

Kl  =  aaaa  .  .  .  .  =  a  taken  n  times  as  a  factor  =  a". 
Substituting  in  (2),  we  obtain 

(a,  +  ay  =  «■  +  noa;"-1  +  '-<"  ~  1]  «V"» 

+  n(n  -  l)(n  -  2)(tV_3  +  §  _  an     .      .    (3) 

I? 
This  formula  is  called  the  Binomial  Theorem;  the  series 
in  the  second  member  is  called  the  expansion  of  (x  +  a)n. 
In  this  expansion  we  observe  the  following 

Rule. 

(1)  The  exponent  of  x  in  the  first  term  is  the  same  as  the 
exponent  of  the  power,  and  decreases  by  unity  in  each  succeed- 
ing term;  the  exponent  of  a  begins  with  one  in  the  second 
term,  and  increases  by  unity  in  each  succeeding  term. 

(2)  The  coefficient  of  the  first  term  is  1,  that  of  the  second 
is  the  exponent  of  the  power,  and  if  the  coefficient  of  any 
term  be  multiplied  by  the  exponent  of  x  in  that  term,  and  the 
product  be  divided  by  the  number  of  the  term,  the  quotient 
will  be  the  coefficient  of  the  next  term. 

By  changing  x  to  a  and  a  to  x,  we  have 

(a  +  x)n  =  a"  +  nan~yx  +  -*— '-  an    W 

If 
+  n(n-  l)fo-2)a„-8a;8+  i<<(f   ,     .(4) 

If  we  write  —a  for  a  in  (3),  we  obtain 
(as  -  a)n  =  x"  -  noaf"1  +  "(w  ~  1}aV-a (5) 

Thus  the  odd  powers  of  a  are  negative  and  the  even  powers 
positive,  and  the  last  term  is  positive  or  negative  according 
as  n  is  even  or  odd. 

Suppose  a  —  1,  then  (4)  becomes 

(l+a0^1+na-  +  M("r1).^+?l(n~1|^~2)^+  .  .  .  *,  (6) 

|2  [3 

which  is  the  simplest  form  of  the  binomial  theorem. 


408  GENERAL    TERM    OF   THE   EXPANSION. 

1.  Expand  (as  +  a)5. 
By  the  rule,  we  have 

(x  +  a)5  =  x5  +  5x4a  +  10x3«2  ■+-  lOara3  +  5xa*  +  a5. 

Similarly 

2.  (a-2o;)7=a7-7«6(2a;)+21a5(2a;)2-3oa4(2x)3+35a3(2.T)4 

-2kr(2x)5+7a(2x)6-(2x)7. 
=a7-14a6x+84a5x2-280a4x3+560a3x4 

-672a2x5+448ax6-128x7. 
Expand  the  following  by  the  Binomial  Theorem : 

3.  (as  -  3)5.     Ans.  x5-  lox4  +  9 Ox3  -  270x2  +  405.x  -  243. 

4.  (3x  +  2?/)4.       81x4  +  21Gx3,y  +  21Gx'-y2  +  9Gxy3  +  1G?/4. 

5.  (x2  +  x)5.  x10 -f5x9  + 10x8-|- 10x7 +  5x9-|-x5. 
G.    (2-fx2)4.  lG-48x2  +  54x4-27x6-hfix8. 

The  sum  of  the  coefficients  in  the  expansion  of  (1  +  a:)"  is  2*.  For 
put  x  =  1 ;  then 

(1  +  .r)»  =  (1  +  1)»  =  2»  =  1  +  n  +  "fr-1)  +  etc. 

=  sum  of  the  coefficients. 
Also,  hy  putting  x  =  —1,  we  have 

(l-l)»  =  l-H  +  "(»-1>-ctc.; 
If 

.".  0  =  sum  of  the  odd  coefficients  —  the  sum  of  the  even  ones; 
i.e.,  the  sums  of  the  odd  and  even  coefficients  are  equal,  and  therefore 
each  =|x2"  =  2n~\ 

191.   The  rth  or  General  Term  of  the  Expansion. 

—  In  the  expansion  of   (x  +  a)",   we  see  that  the  second 

term  is  naJ*-1a ;    the  third  term  is  n'71  ~ — 'x"-'2a-;  and  so 

on  ;  the  last  factor  in  the  denominator  of  each  term  being 
one  less  than  the  number  of  the  term  to  which  it  applies,  one 
greater  than  the  negative  number  in  the  hist  factor  of  the 
numerator,  and  the  same  as   the  exponent  of  a  ;    and  also 


ANY  EXPONENT.  409 

that  the  exponent  of  x  is  found  by  subtracting  the  exponent 
of  a  from  n.     Hence  the 

n(n-  l)(n-  2) (n  -  r  +  2)xn~r +  1ar~1 

1**  term  = |r  —  1 


This  is  called  the  general  term,  because  by  giving  to  r 
different  numerical  values,  any  assigned  term  may  be  ob- 
tained. 

The  coefficient  of  the  rth  term  from  the  beginning  is  equal  to 
the  coefficient  of  the  rth  term  from  the  end. 

The  coefficient  of  the  vth  term  from  the  beginning  is 
n(n  -  1) (n  -  2) O  -  r  +  2) 

\IE± 

By  multiplying  both  terms  by  \n  —  r  +  1,  this  becomes 

^ .     See  (1)  and  (2)  of  Art.  181. 

\r  —  1  \n  —  r  +  1 

The  rth  term  from  the  end  is  the  (n  —  r  +  2)th  term  from 
the  beginning,  and  its  coefficient  is 

*(""  J> r,  which  also  =  1 . 

\n  —  r  +  1  \r  —  1  \n  —  r  +  1 

Therefore  the  coefficients  of  the  latter  half  of  an  expansion 
may  be  taken  from  the  first  half. 

1.  Find  the  fifth  term  of  (a  +  2x3)17. 
Here  n  =  17,  r  =  5  ;  therefore  the 

5th  term  =  17  ' 16  '  W  '  Uals  X  16a;12  =  38080a13^12. 
1-2- 3-4 

2.  Find  the  14th  term  of  (3  -  a)1G.  Ans.  -945a13. 

3.  Find  the    7th  term  of  (a3  +  3a6)9.  61236a1566. 

4.  Find  the    5th  term  of  (a2  -  b2)12.  495a1668. 

5.  Find  the    5th  term  of  (3.^  -  4^)9.  12G  x  36^4V- 

192.  Any  Exponent.  —  To  prove  the  Binomial  Theorem 
when  the  exponent  is  fractional  or  negative. 


410  ANY   EXPONENT. 

Since,  by  Art.  190,  every  binomial  may  be  reduced  to  one 
common  type,  (1  +  %)",  it  will  be  sufficient  to  confine  our 
attention  to  binomials  of  this  form. 

We  bave  seen  that  when  n  is  a  positive  integer, 

(i  +  xy  -. 

We  shall  now  prove  that  this  formula  is  true  when  n  is  frac- 
tional or  negative.  In  this  proof,  which  has  for  its  object  the 
expansion  of  (1  +  x)n  in  the  form  P  +  Qx  -\-  Ax-  +  Bxz  +  etc. , 
we  shall  first  find  P  and  Q,  and  then  determine  the  othei 
coefficients,  A,  B,  etc.,  in  terms  of  P  and  Q. 

(1)    To  find  P  and  Q. 

Since  (1   +  x)"  =  P  +  Qx  +  Ax2  + for  every 

value  of  x,  the  equation  must  be  true  when  x  —  0  ;  therefore 
P  =  1. 

Hence  vl  +  »)■  =  1  +  Qx  +  Ax2  + 


V 
Let  n  be  a  positive  fraction   =  -,  where  p  and   q  are 

positive  integers. 

v 
Assume     (1  +  »)5  =     1  +     Qx  +  etc. 

.-.     (1  +  x)*=  (1  +     Qx  +  etc.)', 
or        1  +  px  +  etc.  =     1  +  qQx  +  etc.  (Art.  11)0). 
.-.     Q  =  |(Art.  185). 
Let  n  be  a  negative  integer  or  fraction  =  —  m  ;  then 

(i  +  .)■  =  (i  + .)-  =  _  2_ 

= =  1  —  iux  +  etc.,  by  division. 

1  +  mx  +  t,,('- 

Hence,   generally,   the   numerical   coefficient  of  a;  in  the 


ANY   KXrONF.NT.  411 

second  term  is  the  same  as  the  exponent,  and  the  form  of 
the  expansion  is 

(1  +  a;)"  =  1  +  nx  +  Ax2  +  Bx3  +  Cx*  + (1) 

(2)    To  find  the  other  coefficients. 

To  lind  A,  B,  etc.,  we  put  x  +  z  for  x  in  (1),  and  then 
expand  1  +  x  +  z  in  two  different  ways  ;  first  regarding 
x  +  z  as  one  term  ;  and  second  regarding  1  -+-  2  as  one 
term.     Thus, 

[1  +  (as  +  2)]"  =  1  +  n(as  +  z)  +  A(x  +  z)*+B(x  +  2)3+  etc. 
=  1  +  nx  +  Ac2  +  Bx8  +  etc. 

+  nz  +  2^1x2  +  o.Ba;2z  +  etc.      .     .     .   (2) 
+  etc.  +  etc. 


[(l+2)+.T]»  =  (l+2)'/l+~Y 


_(l+,)^+_+__+rt8.j 

=  (l+2)"+n(l+2)"-1.«  +  .l(l+2)"-2a;2+etc. 
=  1  +?io;+^lx2+etc. 

+w2  +  w(n  —  l)2x-+yl(?t  —  2)2ar+etc.   .   (3) 

+  etc.  +  etc.  +  etc. 

Now  as  (2)  and  (3)  are  the  expansions  of  the  same 
expression  in  the  same  form,  the  coefficients  of  2,  zx,  zx2, 
etc.,  must  be  the  same  in  both.     Equating  them,  we  get 

n  =  n,     2A  =  n(n  —  1),     oB  =  ^4(n  -  2),     etc. 

.        1_n(n-l)        J_^l(n-2)_n(n-l)(n-2) 

'  [2       '  3        ~  [3  '        ' 

where  the  law  is  obvious.     Substituting  these  values  in  (1), 

we  have 

/1   ■     \.      1            i  n(n—l)    0  ,  n(n  —  l)(n  —  2)    ,  ,  , ... 

(l+a,-)»=l+waH — v  ;s2-j — * ^ ioi3+  etc.,   (4) 

I2  12 

which  proves  the  Binomial  Theorem    for   all  values   of  w, 


412  ANY   EXPONENT. 

whether  fractional  or  negative.     Hence  the  theorem  is  com- 
pletely established. 

Note, — If  n  is  a  positive  integer  in  the  expansion  of  (1  +  x)» 
when  we  form  the  successive  factors,  n  —  1,  n  —  2,  etc.,  the  uth  factor 
will  be  n  —  n,  or  0,  and  therefore  all  the  coefficients  after  that  of  xn 
will  vanish,  for  they  will  all  have  this  factor.  Hence,  the  series  will 
end  with  the  (n  -f  l)th  term.  But  if  n  is  negative  or  fractional,  none 
of  the  factors,  n  —  1,  n  —  2,  etc.,  can  become  0,  and  therefore  the 
series  will  be  infinite. 

1.  Expand  (1  —  a?)«  to  four  terms. 

1    •    /  1    •   Z   •  O 

=  l-fz+iarV^BM- 

2.  Expand  (2  +  o.r)"4  to  four  terms. 

(2  +  3x)-*  =  2-*(l  +  %*)-* 

=  £,1  _«,  +  **_>*,.+ ,. 

The  Binomial  Theorem  may  also  be  applied  to  expand 
expressions  which  contain  more  than  two  terms. 

3.  Expand  (a--2  +  2x  -  l)3. 

Regarding  2x  —  1  as  a  single  term,  the  expansion 

=  (x-2)3  +  3(x-2)2(2.x-  -  1)  +  3x-(2x  -  l)2  +  (2x  -  l)3 
=  x6  +  6a:5  +  9a;4  -  4.x3  -  9a-2  +  6a  -  l,.by  reduction. 

A  root  may  often  be  extracted  by  means  of  the  Binomial 
Theorem. 

4.  Eind  the  cube  root  of  126  to  5  places  of  decimals. 


126^  =  (58  +  1)^  =  5(1  + 


V 


sfi  +  I.I  +  Mi^il1  +S(i-n.(j^)i  + 

V         3     58  1-2     5°  1-2-3         59 


(1+l)». 


EXAMPLES.  413 

Expand  to  1  terras  : 

5.    (1  +  x2)-2.  Am.  1  -  2x2  +  3a;4  -  4x6. 

i  +  x  +  £  _  «!. 

6        54 

7.  Find  0J8  to  5  decimal  places.  9.89949. 

8.  Find  77=  to  5  decimal  places.  0.19842. 


EXAMPLES. 

Expand  by  the  Method  of  Indeterminate  Coefficients  to  4 
terms. 

1.        1  +  2x  o.  ,4ns.  1  +  3a;  +  4a;2  +  7a;3. 

i  +  ix  -  fa;2  +  TV3- 
3  +  fa.  +  ¥aj2  +  2^3 

-.      1  -  ckk  +  a(a  + 1  )x2-  (a3+ 2a2- 1  )xs. 
—  o.»r— x'' 

5.  s/i  + 

Resolve  into  partial  fractions  : 

7x  —  1 


1 

- 

X 

—   X" 

1 

+ 

X 

2 

+ 

X 

+  x- 

3 

+ 

X 

2 

— 

X 

-   X2' 

1 

1+aa;- 

-ax2— x8 

1  _  5X  +  6a;2  1  —  3x       1  -  2x 

-        1  +  3x-  +  2a2  4 3_ 

(1  -  2a;)(l  -  a2)"  1  -  2x       1  -  x 

g            x2  -  10a;  +  13  2                3                4 

(a;  —  l)(a2  —  5a;  +  6)'      a;  —  1        a;  —  2       a;  —  3 

9  9  _J 1 §_ 

(a;  -  l)(aj  +  2)2  x  —  1        a;  +  2        (a;  +  2)2 

in     3a;3 -8a;2 +10  5 7_  1         ,   _3_ 

(a;-l)4     '     (x-iy      (a?-!)8      («-l)a      x-1 


414  EXAMPLES. 


, .                7  -\-  x  .            3        .    4  —  3x 

11. -.  Ans. . 

(1  +  »)(1  +  x1)  1  +  x        1  +  x- 

12            2a;'2  —  lias  +  5  3x                      1 


(x  —  3)  (a;2  +  2x  —  5)  a;2  +  2x  —  5       a;  —  3 

13     a-4  -  3a;3  -  3a;'2  +  10 
(.,  +  i)2(a;  _  3) 

17  11  17 


^l^s.  a;  —  2  + 


16 (a;  +1)        4(a?  +  l)'2        16 (a;  -  3) 
Expand  the  following  by  the  Binomial  Theorem. 

14.  (2a;  -  ?/)5. 

Ans.  32a;5  -  80afy  +  80afy2  -  lO.r?/3  +  lOajy*  -  f. 

15.  (3a  -i-)6. 

Ans.  729a6  -  972a6  +  540a4  -  IGOa3  +  —  -  —  +  — . 


"•c-=y 


A        64a;6   32a-4  ,  20a;2   OA  .  135   243  .  729 
729   27    3        4a;2  Xx*      64a!6 

17.  (1  +  2a;  -  a;2)4. 

Ans.   1  +8a;+20a-'2  +  8a;3-26x4-8a;5+20.t-6-8.i-7-f  .r\ 

18.  (3»2-2aa;+3a2)8. 

Ans.  27a;6-5  loa^+ll  7aV-l  16«V+117(tV--5  la6a;+27o6. 

Write  down  and  simplify  : 

19.  The  4th  term  of  (a;  -  5) 13.  Ans.   -35750a?10. 

20.  The  10th  term  of  (1   -  2a')1'2.  -112640a?. 


22.    The  7th  term 


oil-  +  96  J. 


21.    The  4th  term  of  (     f  9&)   .  40a763. 

10500 


23.    The  5th  term  of  (aAr"<  -  r6"<)8.  70a/y""~^~';- 


EXAMPLES. 


415 


Expand  to  4  terms 

24.  (1  -  x)i. 

25.  (1  -  3®)i 

26.  (1  -  3a,-)~i 

28.  (1  +  fa)"4. 

29.  (8  +  12a)i 

30.  (9  -  6a)-*. 
81.  (4a  -  8x)-K 


flw.S'.   1 


1  —  x  —  a;2  —  fa;3. 
+  x  +  2x2  +  lAe8. 


1  -  x  +  fa;2 


■J  7 


XT'. 


1  _  2a  +  fa'2  -  f«3- 
4(1  +  a  -  £a2  +  |a»). 

AO  +  x  +  f*2  +  fK>- 
2rtiV      «     2  a     2   «/ 


Write  down  and  simplify  : 

32.  The  8th  term  of  (1  +  2x)~$. 

33.  The  5th  term  of  (3a  —  26)"1. 

34.  The  14th  term  of  (210  -  2V)^. 

35.  The  7th  term  of  (38  +  64e)V-. 

Find  to  5  places  of  decimals  the  value  of 


Ans. 


16&4 
243a5' 

-1848a;13. 

^6. 


39.  (630)-".  Ans.  .00795. 

40.  (3128)*.  5.00096. 


36.  V998-     Ans-  9-"332. 

37.  V1003-  10.00999. 
38-  (l^)i  1.00133. 
Expand  to  5  terms  : 

41.  (1  +  3a.-2  -  6a;3)- 1    Ans.  1  -  2a,2  +  4a;3  +  5a;4  -  20a;5. 

42.  (8  -  9a;3  +  18a;4) i .     16(1-  fa;3  +  3a;4  +  &x*  -  -p1) . 


416  RECURRING   SERIES. 


CHAPTER     XX. 

SUMMATION    OF     SERIES. 

RECURRING    SERIES. 

193.  Definition.  —  The  Summation  of  a  Series  is  the 
process  of  finding  an  expression  for  the  sum  of  all  its  terms. 
Examples  of  summation  of  Arithmetic  and  Geometric  series 
have  been  given  in  Chapter  XVII.  We  now  proceed  to  give 
methods  for  summing  other  series. 

A  Recurring  Series  is  one  in  which,  after  a  certain  term, 
each  term  is  equal  to  the  sum  of  a  fixed  number  of  the  pre- 
ceding terms  multiplied  respectively  by  certain  constants. 

A  geometric  progression  is  a  simple  example  of  a  recurring 
series  ;  for  each  term  after  the  first  is  equal  to  the  preceding 
term  multiplied  by  the  ratio.  If  un_1  and  vn  denote  respec- 
tively the  (n  —  l)th  term  and  the  nth  term,  then  ua—  r*/„_,=  0  ; 
the  sura  of  the  coefficients  of  un  and  un_Y  with  their  proper 
signs,  i.e.,  1  —  r,  is  called  the  scale  of  relation. 

In  the  series 

1  +  2x  +  3a2  +  4x3  -f  5.x4  + , 


each  term  after  the  second  is  equal  to  the  sum  of  the  two 
preceding  terms  multiplied  respectively  by  the  constants  2x 
and  —a;2;  these  quantities  being  called  constants  because 
they  remain  the  same  whatever  term  of  the  series  we  consider. 
Thus, 

5x*  =  2a-4<B8  +  (-^)  •••>•<•"; 

that  is,  wn  =  2a;  u„_!  —  x2  u„_», 

or  un  —  2xu„_i  +  x"u„_2  =  0. 

This  law  holds  for  all  values  of  n  greater  than   1,  so  that 


SCALE   OF  RELATION   GIVEN   TO  FIND   ANY  TERM.    417 

every  term  after  the  second  can  be  obtained  from  the  two 
terms  immediately  preceding. 

Thus  the  series  1  +  2x  +  ox2  +  4x3  +  5x*  + 

is  a  recurring  series  in  which  the  scale  of  relation  is  1  —  2x+x2. 

194.  The  Scale  of  Relation  being  Given  to  Finfl 
Any  Term.  —  If  the  scale  of  relation  of  a  recurring  series 
is  given,  any  term  can  be  found  when  a  sufficient  number  of 
the  preceding  terms  are  known.  As  the  method  of  proceed- 
ing is  the  same  however  many  terms  the  scale  of  relation 
may  have,  the  following  illustration  will  suffice. 

If  1  —  px  —  qx2  —  rxs  is  the  scale  of  relation  of  the  series 

a0  +  axx  +  a2x2  -+-  a3x3  + , 

we  have  anxn  =  px  •  a^x"-1  +  qx2  ■  an_2xn~2  +  ra?  •  a^x"'3, 

or  a„  =  2>a0_i  +  q^n-i  +  ran-z'i (1) 

thus  any  coefficient  can  be  found  when  the  coefficients  of  the 
three  preceding  terms  are  known. 

Conversely.  To  find  the  scale  of  relation  when  we  have 
given  a  sufficient  number  of  the  terms  of  a  series. 

Find  the  scale  of  relation  of  the  recurring  series 

2  +  4a;  +  14a-2  +  46a;3  + 

Let  the  scale  of  relation  be  1  —  })x  —  qx2 ;  then  to  find  j? 
and  q  we  have  from  (1)  the  equations, 

14  -  Ap  —  2q  =  0,  ami  46  —  14p  -  4q  =  0  ; 

whence  p  =  3,  and  q  =  1  ;  thus  the  scale  of  relation  is 

1  -  3aj  -  x2. 

If  the  scale  of  relation  consists  of  3  terms,  it  involves  2  constants, 
p  and  q ;  and  therefore  we  must  have  2  equations.  To  obtain  p  we 
must  have  given  at  least  3  terms  of  the  series,  and  to  obtain  q  we  must 
have  one  more  term  given.  Thus,  to  obtain  a  scale  of  relation  involv- 
ing 2  constants,  we  must  have  given  at  least  4  terms  of  the  series. 

Similarly,  if  the  scale  of  relation  be  1  —  px,  —  qx2  —  rx3,  we  must 


418  TERMS    OF  A    RECURRING    SERIES. 

have  X  equations  to  find  the  '->  constants,  and  therefore  at  least  (5  terms 
of  the  series  must  he  given. 

In  general,  if  the  scale  of  relation  involve  m  constants,  we  must 
have  given  at  least  2m,  consecutive  terms. 

Conversely,  if  2m  consecutive  terms  are  given,  we  may  assume  for 
the  scale  of  relation  1  —  pxx  —  p2x2  — —  pm-tm. 

195.  To  Find  the  Sum  of  n  Terms  of  a  Recurring 
Series.  —  The  method  of  finding  the  sum  is  the  same  what- 
ever be  the  scale  of  relation  ;  for  simplicity  we  shall  suppose 
the  scale  to  contain  only  two  constants. 

Let  the  series  be  a0  -f-  a^  +  a.,x'2  -f-  asxz  + ,  and  let 

the  scale  of  relation  be  1  —  px  —  qx2,  so  that  for  every  value 
of  n  greater  than  1  we  have  an  —  pan_x  —  qan-2  =  0. 
Denote  the  sum  by  S ;  then 

S  =s  a0  +  atx  +  a»x2  + +  an_lxn~l. 

—  p<K-^~l  -F'„-i'-n. 

— qx2S  =  —qa0x2  —  ...  —  qan  _3xn~1—  qan  _  2xn  —  qa„ ..  x  a?  K  ] . 

.*.     (1—  px— qx2)S=a0  +  (a1  —  pa0)x— (pan-i+qan-2)xn  —  qa„-i.rn  +  ', 

for  all  the  other  terms  in  the  second  member  disappear  in 
consequence  of  the  relation  an  —  pan_j  —  qan_2  —  0. 

.      s  =a0+  (>i—  pa^x—  [(pa„-l  +  qa„_2)xn+qan_lxn+1~\     ... 
1  —  px  —  qx2 

Thus,  the  sum  of  a  recurring  series  is  a  fraction  whose 
denominator  is  the  scale  of  relation. 

If  the  term  \_{pan_x  +  qan-'>)xn  -f  gam_iaf +1]  decreases 
indefinitely  as  n  increases  indefinitely,  we  have  for  the  sum 
of  an  infinite  number  of  terms, 

s  _  a„  +  («.,-  pa0)x .gv 

1  —  px  —  qx* 

If  this  fraction  be  developed  into  a  series  according  to  the 
ascending  powers  of  .<•.  we  shall  obtain  the  given  recurring 
series  (Art.  180)  ;  for  this  reason  the  fraction  is  called  the 


SERIES  SUMMED   BY  MEANS   OF   OTHER  SERIES.    419 

generating  fraction*  of  the  series.  Thus,  when  the  series 
is  convergent,  the  summation  reproduces  the  generating 
fraction. 

Find  the  generating  fraction  of  the  following  series : 

1.    4  +  9<e  +  21a2  +  51a3  + 

To  find  2}  and  q,  we  have 

21  =  dp  +  4g,  and  51  =  21p  +  9^ ; 
whence  p  —  5,  and  q  =  —  6. 

Substituting  in  (2),  we  get 

_  4  +  (9  -  20)a  _       4  -  11a; 
~~    1  —  ox-  +  6a2 


1  - 

ox  +  6a;2 

Ans.         1+X       . 
l-10a+21a2 

1  +  3a 

(1  -  xy 

1  +  hx 

1  -  x  -  6a;2 

2.  l  +  lla+89a2  +  659a3+  .  . 

3.  1  +  5a  +  9a-2  +  13a-8  + 

4.  1  +  6a  +  12a2  +  48a3  + 

196.   Series  Summed  by  Means  of  Other  Series. 

—  When  a  series  is  the  difference  between  two  other  simpler 
series  it  may  often  be  summed  by  taking  the  difference  of 
the  two  latter  series.  Even  though  the  sums  of  these  two 
series  are  not  knowu,  the  difference  of  these  sums  can  be 
found  when,  after  a  certain  number  of  terms,  the  succeeding 
terms  of  the  two  series  are  identical. 

1.    Find  the  sum  of  n  terms  of  the  series, 

«   +-L  +  A+  ! 


1-2        2-3        3-4  n(n  +  1) 

Each  term  of  this  series  may  be  divided  into  two  parts ; 

thus, 

J_  =  1  _  1        1     =1       1  1        =  1 1_  - 

1-2       1       2'   2-3      2      31 n(w  +  l)      n      n+1 ' 

*  Called  also  generating  function. 


420     SERIES  SUMMED   BY   MEANS   OF   OTHER  SERIES. 

Hence  the  sum  =  U  -  A)  +  (A  -  A)  +  .  .  .  +  (l —\ 

\n      n+lj 


i-    x 


n  +  1       w  +  1 

for  all  the  other  terms  disappear,  since  the  second  part  of 
each  term,  except  the  last,  is  canceled  by  the  first  part 
of  the  next  succeeding  term. 

When  n  is  infinite,   becomes  zero,  and  the  sum 

n  +  1 
equals  1. 

Note.  —  The  expression  for  the  nth  term  of  a  series  in  terms  of  n  is 
called  the  general  term  of  the  series.  It  is  obvious  that  when  the  nth 
term  of  a  series  is  given  we  can  write  down  all  the  terms  by  simply 

substituting  J,  2,  3, successively  for  n.     It  is  not  true, 

however,  that  when  the  first  few  terms  are  given  we  can  in  general 
find  the  nth  term,  if  nothing  is  told  us  regarding  the  form  of  that  term. 

2.    Find  the  sum  of  n  terms  of  the  series, 

+  JL  +  JL  +  A  + 


1.4       2-5       3-6       4-7  .  n(n  +  3) 

Each  term  may  be  written  in  the  form 

*(*-*)+*<*-*)+*<*  -*)+ +  i(u~^b} 

.-.    Sum  =  A^l  +  A  +  A  +  J  +  i  + +  £ 

_i_!_        _i__i i L_l 

4      5       n      n+1      n  +  2      n  +  3 J 

_  iA   -  i  .  i i i L_\ 

~  X  +  S  +  3       n  +  1       n  +  2       n  +  3/ 

since  all  the  terms  disappear  by  canceling  except  three  nt 
the  beginning  and  three  at  the  end. 

When  n  is  infinite  the  sum  becomes  {'. 

Find  the  sum  of  the  following  series,  (1)  to  n  terms,  and 
(2)  to  infinity : 


METHOD    OF  DIFFERENCES.  421 


s.  JL  +  JL  +  JL  + 


1-8       3-5       5-7  (2«  -  l)(2w  +  1) 

Ans.  (l)r-^-T,  (2)f 
2?i  +  1 

4.    J_   +  A.  +  J_  + +  1         . 

1-8       2-4       3-5  w(n  +  2) 

Ans.  (1)  3?t'2  +  5w ,  (2)1. 

v      4(w2  +  3n  +  2)    v   '4 

METHOD    OF    DIFFERENCES. 

197.  Definition  —  To  Find  Any  Term  of  a  Series. 

—  When  we  have  a  series  whose  terms  proceed  according  to 
any  law,  if  we  take  the  first  term  from  the  second,  the  second 
from  the  third,  the  third  from  the  fourth,  and  so  on,  the 
several  remainders  will  form  a  new  series  called  the  first 
order  of  differences. 

If  the  differences  of  the  terms  of  this  new  series  be  taken 
in  the  same  manner,  we  shall  obtain  another  series  called  the 
second  order  of  differences;  and  so  on. 

Let  a,  b,  c,   d,  e,  f, be  the  series ;    then  the 

successive  orders  of  differences  are, 

1st  order,  b  —  a,     c  —  b,     d  —  c,     e  —  d,    /  —  e,  .  .  .  . 

2d  order,  c  —  2b  +  a,     d  —  2c  +  6,     e  —  2d  +  c,  .  .  .  . 

3d  order,  d  —  3c  +  '3b  —  a,     e  —  3d  +  3c  —  6,  .  .  .  . 

4th  order,  e  —  4d  +  6c  —  4b  +  <*»•••• 
and  so  on,  where  each  difference,  though  a  compound  quan- 
tity, is  called  a  term. 

If  we  let  du  d2,  d3,  d4, represent  the  first  terms  of 

the  first,  second,  third,  etc.  orders  of  differences,  we  shall 
have 

di  =  b  —  a,  .*.     6=a+  dx. 

d2  =  c—2b+a,  .'.     c  =  a  +  2dx+  d2. 

ds  =  d-3c+3b— a,  .-.     d=a-\-3d1-\-3d2-\-  ds. 

dt  =  e— 4d+6c— 46+a,  .-.  e==  a  +  4cZ1+6d2+4cZ3+d4. 
etc.  =  etc.  etc.  =  etc. 


422  TO   FIND   ANY   TERM    OF  A   SERIES. 

We  see  here  that  the  coefficients  of  the  value  of  any  term, 
as  c  for  instance,  the  third  term  of  the  series,  are  the  Bame  as 
the  coefficients  of  the  second  power  of  a  binomial ;  the  coeffi- 
cients of  the  value  of  cZ,  the  fourth  term,  are  the  same  as  the 
coefficients  of  the  third  power  of  a  binomial ;  and  so  on. 
Hence  we  infer  that  the  coefficients  of  the  ?tth  term  of  the 
series  are  the  same  as  the  coefficients  of  the  («  —  l)"1  power 
of  a  binomial.*  Thus  the  nth  term  of  the  series  a,  b,  c,  d, 
e, is 

a-Kn-l)^^-1)^"-2)^1-1)^2^"-3^.  •  •   (1) 

LA  IA 

1.  Find  the  10th  term  of  the  series  12,  40,  90,  168,  280, 
432, 

The  successive  orders  of  differences  are 

1st  order,        28,  50,  78,  112,  152, 

2d  order,  22,  28,    34,    40, 

3d  order,  6,       6,      6, 

4th  order,  0,      0, 

Then  a  =  12,  d,  =  28,  d2  =  22,  d,  =  6,  dt .  .  .  =  0,  n=  10. 
Hence  from  (1)  we  have  10th  term 

=  12  +  9.28+^22  +  ^-^6=12  +  252  +  792  +  504=1560. 

2.  Find  the  12th  term  of  the  series  2,  6,  12,  20,  30,  .  .  . 

Ans.  156. 

3.  Find  the  nth  term  of  the  series  1,  3,  6,  10,  15,  21,  .  .  . 

Ans.  — -C — . 
■2 

198.   To  Find  the  Sum  of  n  Terms  of  any  Series. 

—  Let  the  given  series  be 

a,  6,  c,  d,  e,  /, (1) 

and  denote  the  sum  of  the  first  n  terms  by  S. 

*  For  a  proof  of  this  by  Induction,  see  Sail  and  Knight's  Higher  Algebra,  p.  324. 


TO   FIND    THE   SUM    OF  n    TERMS    OF  ANY   SERIES.    423 

Assume  the  series 

0,  a,  a  +  &,  a  +  &  +  c,  a  +  6  +  c  +  dz (2) 

It  is  clear  that  the  sum  of  n  terms  of  the  given  series  is 
equal  to  the  (n  +  l)th  term  of  the  assumed  series  ;  also  the 
first  order  of  differences  of  series  (2)  is  the  same  as  series 
(1);  hence  the  second  order  of  differences  of  series  (2)  is 
the  same  as  the  first  order  of  series  (1);  the  third  order  of 
series  (2)  is  the  same  as  the  second  order  of  (1);  and  so 
on. 

Hence  by  (1)  of  Art.  197,  we  may  obtain  the  value  of  S 
by  putting  a  =  0,  n  =  n  +  1,  clx  =  a,  tl,  =  d1:  ds  =  d2, 
etc.     Making  these  substitutions,  we  have 

S  =  na  +  "C"-1)^  +  H(n-l)(»-2)<t[  +  (8) 

Note.  —  It  will  be  seen  that  this  method  of  summation  is  exact 
only  when  the  series  is  such  that  in  forming  the  orders  of  differ- 
ences, we  eventually  come  to  a  series  in  which  all  the  terms  are  equal. 
When  there  are  no  orders  of  differences  whose  terms  are  equal,  we 
may  obtain  by  (3)  approximate  results,  which  will  be  nearer  the  truth 
the  greater  the  number  of  terms  used. 

1.  Find  the  sum  of  n  terms  of  the  series  1,  3,  5,  7,  9,  .  .  . 
Here  a  =  1,  dj  =  2,  d2, =  0.     Hence  from   (3) 

wg  have 

g  =  n  +  n(n  7  ^  2  =  w2.      (See  Ex.  4  of  Art.  161). 

Find  the  sum  of  the  following  series  : 

2.  1,  3,  6,  10,  15,  21, ton  terms. 

Ans.  n(n  +  l)(n  +  2)> 
6 

3.  1  •  2,  2  •  3,  3  •  4,  4  •  5,  5  •  G, tow  terms. 

A       w(n +  !)(»  + 2) 
3 

4.  3,  11,  31,  G9,  131, to  20  terms.  .     44330. 


424 


P1LE8    OF   CAX.XOX-BALLS. 


199.  Piles  of  Cannon-Bails.  —  Cannon-balls  are  usually 
piled  ou  a  level  surface  in  horizontal  courses,  in  piles  of  three 
different  forms,  triangular  pyramids,  square  pyramids,  and 
rectangular  piles. 


TRIANGULAR    PYRAMID. 


SQUARE    PYRAMID 


(1)  In  a  triangular  pyramid  the  summit  consists  of  1  ball, 
the  second  course  consists  of  3  balls,  the  third  course  con- 
sists of  6  balls,  the  fourth  of  10  balls,  and  so  on.  Hence 
the  number  of  balls  in  a  triangular  pile  of  n  courses  is  equal 
id  the  sum  of  n  terms  of  the  series, 

1,  3,  G,  10,  15, 

But  from  Ex.  2  of  Art.  198,  the  sum  of  n  terms  of  this 


^n(n  +  l)(n  +  2) 
6 


(1) 


in  which  n  is  the  number  of  balls  in  the  side  of  the  bottom 
course,  as  well  as  the  number  of  courses.  % 

(2)  In  a  square  pyramid  the  summit  consists  of  1  ball, 
the  second  course  consists  of  4  balls,  the  third  course  consists 
of  9  balls,  and  so  on.  Hence  the  number  of  balls  in  a 
square  pile  of  n  courses  is  equal  to  the  sum  of  n  terms  of 

the  series 

l2,  22,  32,  42, n\ 

Here  a  —  1,  d\  =  3,  da  =  2,  cls,  .  .  .  .  =  0.  Hence  by 
(3)  of  Art.  198, 


S=n  + 


Sn(n 


1)   ,  n(n- 


l)(n-2)  _»(H  +  l)(2n  +  l) 


(2) 


2  '  3  6 

Here  also  n,  is  the  number  of  'mils  in  the  side  of  the  bottom 
eoursevas  well  :is  the  cumber  of  courses. 


PILES   OF  CANNON-BALLS.  425 

(8)  In  a  rectangular  pile  the  summit  consists  of  1  row, 
the  second  course  consists  of  2  rows,  the  third  of  3  rows,  and 
so  on.  Let  m  +  1  denote  the  number  of  balls  in  the  top 
row  ;  then  the  length  of  the  second  row  will  be  m  +  2,  of 
the  third  m  +  3,  and  so  on.  Hence  the  summit  will  consist 
of  m  +  1  balls,  the  second  course  will  consist  of  2(m  -+■  2) 
balls,  the  third  course  of  3(m  +  3)  balls  ;  and  the  nth  course 
of  n(m  +  n)  balls.  Therefore  the  number  of  balls  in  the 
rectangular  pile  of  n  courses  is  equal  to  the  sum  of  n  terms 
of  the  series 

m  +  1,  2(ra  +  2),  3(m  +  3), n(m  +  n). 

Here  a  =  m  +  1,  c7x  =  m  +  3,  d2  =  2,  d3, =  0. 

Hence  by  (3)  of  Art.  198, 

S  =  n(m  +  1)  +  n(n~1}(m  +  3)  +  "("  -  l)(n  -  2) 
&  o 

_  n(n  +  l)(3m  +  2n  +  1) 

6 

If  I  =  the  number  of  balls  in  the  longest  row,  then 
I  =  m  +  n,  and  therefore  3m  +  2n  =  ol  —  w,  and  the 
formula  becomes 

<y  _  n(n  +  1)(3?  -  n  +  1)  .   . 

^>  -  g  ,  ....    (3) 

in  which  Z  denotes  the  number  in  the  length,  and  n  the  num- 
ber in  the  width,  of  the  bottom  course. 

To  find  the  number  of  balls  in  an  incomplete  pile,  we  must 
find  the  number  in  the  pile  supposed  complete,  then  the 
number  in  the  pile  which  is  wanting,  and  subtract  the  latter 
from  the  former. 

1.  Find  the  number  of  balls  in  a  square  pile  of  15  courses. 

Ans.  1240. 

2.  Find  the  number  of  balls  in  a  rectangular  pile,  the 
length  and  breadth  of  the  bottom  course  containing  GO  and 
30  balls  respectively.  Ans.  23405. 


426  INTERPOLATION. 

3.  How  many  balls  in  an  incomplete  triangular  pile,  the 
number  of  balls  in  each  side  of  the  lower  course  being  44, 
and  in  each  side  of  the  upper  22  ?  Ana.    13409. 

200.  Interpolation.  —  The  process  by  which  we  intro- 
duce between  terms  of  a  series,  intermediate  terms  which 
conform  to  the  law  of  the  series,  is  called  Interpolation.  Its 
most  extensive  application  is  in  Astronomy,  though  it  is 
often  used  for  inserting  intermediate  terms  between  those 
given  in  Logarithmic  and  other  Mathematical  tables.  To 
insert  or  interpolate  a  term  in  a  series,  is  practically  to  find 
the  nth  term  of  the  series  by  the  method  of  differences  (Art. 
197). 

Let  p  represent  the  distance  of  the  required  term  from  a, 
the  first  term  of  the  series,  this  distance  being  measured  in 
terms  of  the  distance  between  any  two  consecutive  terms. 
Hence  p  will  be  a  fraction  ;  and  since  it  is  the  distance  from 
a  to  the  ntb  term  of  the  series,  it  is  the  distance  to  the 
(n  —  l)"1  term  from  a;  therefore^  =  n  —  1.  Substituting 
in  (1)  of  Art.  197,  we  obtain  for  the  term  to  be  interpolated 

a  +  M  +  S&^d,  +  pCp  ~  'Hp  -  *>rf,  +   .  (1) 

1.  The  cube  roots  of  GO,  Gl,  62  are  respectively  3.91487, 
3.93650,  3.95789  :  find  the  cube  root  of  60* 


1st  order  of  differences  =  0.02163,  0.02139, 
2d  order  of  differences  =  -0.00024,  .... 


.-.     a  =  3.91487,  d,  =  0.02163,  d,  =  -0.00024,  j>  =  \. 
Substituting  in  (1),  we  have  for  the  required  term 

3.91487  +  K.02163)  +  j  (-f)/-  -0"024) 

=  3.91487  +  .00541  +  .00002  =  8.92030. 


REVERSION    OF  SERIES.  427 

2.  The  logarithm  of  radius  vector  of  Mars 

1888,  July    7  is  0.1782987. 

"     11  is  0.1767934. 

«        "     15  is  0.1752860. 

"     19  is  0.1737790. 

What  is  it  (1)  for  July  8th  ;  and  (2)  for  July  10th? 

Ans.  (1)  0.1779227;  (2)  0.1771700. 

3.  Given   V^  =  2.23607,  \[H  =  2.44949,    V^  =  2.64575, 
vAs  =  2.82843  :  find  ^5725,  V^,  and  s/~5J). 

Ans.  2.29129,  2.32379,  2.34521. 

4.  Given  V49  =  3.65931,  y/51  =  3.70843,  V53  =  3.75629, 

V55  =  3.80295  :  find  V50,  V^  and  V5^- 

Ans.  3.68403,  3.73251,  3.77976. 

201.   Reversion  of  Series.  —  To  Revert  a  Series  is  to 

express  the  value  of  the  unknown  quantity  contained  in  the 

series  by  means  of  another  series  involving  the  powers  of 

some  other  quantity.     Thus,  given  the  series 

1.  y  =  ax  +  bx2  +  ex3  -f- (1) 

then  to  revert  this  series  is  to  express  the  value  of   x  in 
another  series  involving  the  powers  of  y. 

Assume       x  =  Ay  +  By2  +  Cy3  + (2) 

Substituting  in  (2)  the  value  of  y  from  (1),  we  have 

x  =  Aax  +  Abx2  +  Acx3  + 

+  Ba2x2  +  2Babx3  ■+-  .  .  .  . 
+  Ca3x3  + 


Therefore  (Art.  185) ,  Aa  —  1 ; 

...  4  =  1. 

a 

Ab  +  Ba2  =  0  ; 

a3 

Ac  +  2Bab  +  Co3  =  0 ; 

„       262  -  ac 

.  .     u  =  ■                 '. 
a5 

which  in  (2)  gives 

a        a3                 a0 

«c).v3  _^ 

428  EXAMPLES. 

When  the  given  series  contains  only  the  odd  powers  of  x, 
we  may  shorten  the  process  by  assuming  for  as  another  series 
containing  only  the  odd  powers  of  y.     Thus, 

2.  Revert  the  series  y  =  ox  +  bx*  +  ex5  + 

Here  we  assume  x  =  Ay  +  By3  +  Cy5  + 

If  we  assumed  as  we  did  in  Ex.  1,  the  alternate  coefficients 
would  each  reduce  to  zero.    Solving  this  example,  we  obtain 
y        bf        (3b2  -  ctc)yb 
x  =  a  ~   a4  +  a7 

If  the  first  term  of  the  series  to  be  reverted  in  any  case 
be  independent  of  x,  thus  if  the  series  be  of  the  form 
y  =  a'  +  ax  +  bx2  +  etc.,  we  must  put  y  —  a'  =  z,  and 
express  x  iu  a  series  involving  the  powers  of  z,  and  /hen 
replace  the  value  of  z. 

Revert  the  following  series  : 

3.  y  =  X  +  X2  +  x*  4- 

Ans.  x  =  y  —  y2  +  y3  —  y*  + 

4.  V  =  x  +  -  +  |-  + 

?/2   .    ?/3         ?/4    . 

^•*  =  y      2+JA~Ll+ 

5.  y  =  x  -  2x2  +  3.x3  - 

Ans.  x  =  y  +  2y2  +  5?/3  +  Uy*  + 


EXAMPLES. 

Sum  the  following  ten  recurring  series  to  infinity : 

1  —  2aj 
1.1+  3a;  +  11a;-  +  43x3  +  .  .  .  .     -4ns.        .  . 

2.  1  +  2k  +  3z2  +  4z3  +  5a*1  + _        • 

3.  !  +  2,  +  8*2  +  28.,3  + r^l^^^ 

4.  2  -  x  +  5ar»  -  7*3  + 1  +  aT-V 


EXAMPLES.  429 

7  20x 

5.  7  -  6x  +  9x'  +  27z«  + Am.  t  _  2a.  _  ^ 

6.  1  +  3s  +  5*  +  1*  + T=£t*- 


7.  1  +  Sx  +  4s«  +  '*■  + 

8.  2  +  3X  +  W  +  M+ f_ 


1  +  2x 
—  x  - 
2  -  3x 

9.    3  _j_  6a;  +  14a;2  4-  36a;3  +  98a;4  +  276a;5  +  .....'. 
A  3  -  12a;  +  H«a 

*  '  1  —  6a;  +  11a;2  —  6a;3' 

10.  2  -  a;  4-  2a;2  -  5a;3  4-  10a;4  -  Ux5  4- 

2  +  5a;  +  5x\ 

(1  +  *)■ 

Sum  the  following  six  series  : 

11.  i+ J- +  J- +7^  +  ...  to  infinity.    Am.  f  f. 

l-o       2  •  b       3-7       4  • « 

12     _i I ■  H 1- (1)  to  n  terms,  and 

'    8-8^6.12       9-16 

(2)  to  infinity.     [Multiply  by  3  •  4.] 

13.  _1 1 1 1 to  infinity.  J. 

1.3      2-4       3-5 

14.  — 1 1 h (1)  to  n  terms,  and  (2) 

3-4      4-5       5-6 

to  infinity.  Ans.  (1)  g      n+       ;     (2)  f 

15.  —  +  —  +  —  + (1)  to  n  terms,  and  (2) 

1  . 3       2-4       3-5 

to  infinity.  Ans.  (1)  f(f  -  nJ+   +  +  2)  5     (2)  *■ 

16.  — — I — I 1 +  ...  to  infinity.       1. 

1.5^5.9^9.13^13.17  T  J 


430  EXAMPLES. 

Find  the  nth  term  and  the  sum  of  n  terms  of  the  series : 

17.  1,  4,  10,  20,  35, 

Ans   n(n+  l)(w  +  2)  ,  71(71  +  l)(n  +  2)(n  +  3) 
[8  [± 

18.  4,14,30,52,80,114, 3w9  +  »;  n(n  +  l)2. 

10.    8,  26,  54,  92,  140,  198, 

Aits.  5n2  -f  3n  ;  £w(n  +  1)(5»  +  7). 

20.  2,  12,  36,  80,  150,  252, 

Ans.  n\n  +  1)  ;  TV*0  +  l)(w  +  2)(3n  +  1). 

21.  8,  16,  0,  -64,  -200,  -432,  - 

Aris.  — 4n2(?i  —  3)  ;   —  n(n  +  l)(n2  —  3?i  —  2). 

22.  30,  144,  420,  960,  1890,  3360, 

Ans.  n(n+l)  (n+2)  (w+4)  ;  ^-(w+l)(w+2)(n+3)(4w+21). 

Find  the  number  of  balls  in 

23.  A  triangular  pile  of  18  courses.  Ans.  1140. 

24.  A  rectangular  pile,  the  length  and  breadth  of  the 
base  containing  50  and  28  balls  respectively.        A)is.  16646. 

25.  An  incomplete  triangular  pile,  a  side  of  the  base 
having  25  balls,  and  a  side  of  the  top  14.  A71S.  2470. 

26.  An  incomplete  square  pile  of  27  courses,  having  40 
balls  in  each  side  of  the  base.  Ans.  21321. 

27.  An  incomplete  square  pile  having  169  balls  in  the  top 
course,  and  1089  in  the  lowest  course.  Ans.  11879. 

28.  A  complete  rectangular  pile  of  15  courses,  having  20 
balls  in  the  longer  side  of  its  base.  Ans.  1840. 

29.  An  incomplete  rectangular  pile,  the  number  of  balls 
in  the  sides  of  its  upper  course  being  11  and  18,  and  the 
number  in  the  shorter  side  of  its  lowest  course  being  30. 

Ans.  11940. 

30.  An  incomplete  square  pile  of  16  courses  when  the 
number  of  balls  in  the  upper  course  is  1005  less  than  in 
(he  lowest  course.  Ans.  L8296. 


EXAMPLES.  431 

31.  The  number  of  balls  in  a  complete  rectangular  pile  is 
24395  ;  if  there  are  34  balls  in  the  breadth  of  the  base,  how 
many  are  there  in  its  length?  A>is.  52. 

32.  The  number  of  balls  in  a  triangular  pile  is  greater  by 
150  than  half  the  number  of  balls  in  a  square  pile,  the 
number  of  courses  in  each  being  the  same :  find  the  number 
of  balls  in  the  lowest  course  of  the  triangular  pile. 

Arts.  300. 

33.  Show  that  the  number  of  balls  in  a  square  pile  is  one- 
fourth  the  number  of  balls  in  a  triangular  pile  of  double  the 
number  of  courses. 

34.  If  the  number  of  balls  in  a  triangular  pile  is  to  the 
number  of  balls  in  a  square  pile  of  double  the  number  of 
courses  as  13  to  175,  find  the  number  of  balls  in  each  pile. 

Ans.  Triangular  364  ;  Square  4900. 

35.  The  square  roots  of  30,  31,  32,  33  are  respectively 
5.47723,  5.56776,  5.65685,  5.74456  :  interpolate  the  square 
roots  of  30.3,  30.5,  30.8.      Ans.  5.50454,  5.52268,  5.54978. 

36.  The  cube  roots  of  9,  11,  13,  15  are  respectively 
2.08008,  2.22398,  2.35133,  2.46621  :  interpolate  the  cube 
roots  of  10,  12,  14.  Ans.  2.15443,  2.28943,  2.41014. 

37.  The  reciprocals  of  74,  75,  76,  77  are  respectively 
.01351,  .01333,  .01316,  .01299:  interpolate  the  reciprocals 
of  74.2,  74.4,  74.7.  Ans.  .01348,  .01344,  .01339. 

38.  The  logarithms  of  232,  233,  234,  235  are  respectively 
2.365488,  2.367356,  2.369216,  2.371068:  interpolate  the 
logarithms  of  233.4,  233.6,  233.8. 

Ans.  2.368101,  2.368473,  2.368845. 

39.  The  right  ascension  of  Jupiter  on  a  certain  Monday 
at  noon  was  10  h.  5  m.  38.6  s.  ;  on  Tuesday  at  noon  it  was 
10  h.  6  m.  18.86  s.  ;  on  Wednesday  at  noon  10  h.  6  m. 
59.41  s.  ;  on  Thursday  at  noon  10  h.  7  m.  40.24  s.  Find 
its  right  ascension  on  Tuesday  at  midnight. 

Ans.  10  h.  6  m.  39.1  s. 

40.  The  latitude  of  the  moon  1888,  March  26,  Monday,  at 
noon  was  3°  9'  34."4  ;  at  midnight  3°  38'  42. "5  ;  on  Tuesday 


432  EXAMPLES. 

at  noon  4°  4'  23."5  ;  at  midnight  4°  20'  l/'l  ;  on  Wednesday 
at  noon  4°  43'  4. "4.     Find  its  latitude  at  !)  p.m.  on  Tuesday. 

Ans.  4°  21/  l."3. 
Revert  the  followiug  series  : 

41.    y  =  x  +  3-«'2  +  oar5  +  7x4  + 

Ans.  x  =  y  -  3/  +  13//3  -  b^f  + 

42.,  =  *-f  +  f-...*  =  ,+  |°  +  £  +  |+... 

*,  —  f +  ?-*+-..... 

Ans.  x  =  y  +  £+  ^f  +  ^if  + 

44.    y  =  2x  +  Sx3  +  4.r5  +  5x7  + 

Ans.  x  =  \y  -  T%?/3  +  TV9g.V5  - 


45.    y  =  x  +  f  +  f+...      x  =  y-£  +  y~-£  + 


2        6  2       3       4 


46.    ?/  =  1  —  2x  +  3ar  — 

Ans.  x=-  i0/  -  1)+  10/  -  I)2  -  MV  ~  O8  + 


47.    y  =  1  +  ^  +  t  +  f-  + 
2  b 


^ws.  a?  =  (y  -  1)  -  K^  -  !)2  +  1(2/  ~  !)8  ~  •  • 

48.    Given  y  +  a?/'2  +  ^3  +  .  .  .  =  px  +  go:2  +  rx3  +  , 
to  express  x  in  terms  of  y. 
Ans.  x=V-+  («P2-?).V2  +  R>j>4-pr-2g(ap2-r7)]?/3  |    _ 


_p  p«  p- 


LOGARITHMS  —  DEFINITIONS.  433 


CHAPTER    XXI. 

LOGARITHMS— EXPONENTIAL   AND    LOGARITHMIC 
SERIES  — INTEREST    AND    ANNUITIES. 

LOGARITHMS. 

202.  Definitions.  —  The  Logarithm  of  a  number  is  the 
exponent  of  the  power  to  which  another  number,  called  the 
base,  must  be  raised  to  equal  the  given  number. 

Thus,  if  ax  =  N,  x  is  called  the  logarithm  of  iVto  the  base 
a.  This  is  usually  written  logaiV,  so  that  the  same  meaning 
is  expressed  by  the  two  equations 

ax  =  N;  x  =  logaJV. 

Examples.  (1)  Since  34  =  81,  4  is  the  logarithm  of  81 
to  the  base  3. 

(2)   Since  101  =  10,  102  =  100,  103  =  1000, the 

natural  numbers  1,2,  3, are  respectively  the  loga- 
rithms of  10,  100,  1000, to  the  base  10. 

A  System  of  Logarithms  means  the  logarithms  of  all  positive 
numbers  to  a  given  base.  Thus,  if  we  suppose  a  to  remain 
fixed  while  N  assumes  in  succession  every  value  from  0  to  oo, 
the  corresponding  values  of  x  will  constitute  a  system  of 
logarithms  to  the  base  a.  The  base  is  then  called  the  base 
of  the  system.  Any  number  might  be  taken  as  the  base  of 
the  system,  and  corresponding  to  any  such  base  a  system 
of  logarithms  of  all  numbers  could  be  found. 

Thus,  suppose  a  =  9  ;  then 

Since  91  =       9,         1  =  log9      9. 

Since  9'=    81,         2  =  log9    81. 

Since  93  =  729,         3  =  log9  729. 

That  is,  the  natural  numbers  1,  2,  3,  .  .  .  .  are  respectively 


434  PROPERTIES   OF  LOGARITHMS. 

the  logarithms  of  9,  81,  729,  ....  in  the  system  whose  base 
is  9. 

Note.  —  When  it  is  understood  that  a  particular  system  of  loga- 
rithms is  in  use,  the  suffix  denoting  the  base  is  omitted.  Thus  in 
Arithmetic  calculations  in  which  10  is  the  base,  we  usually  write  log 
2,  log  3, instead  of  log)02,  log,  03, 

203.  Properties  of  Logarithms.  —  Before  discussing 
the  logarithmic  systems  in  common  use,  we  shall  prove  some 
general  Properties  which  are  true  for  all  logarithms,  whatever 
the  base  may  be. 

(1)  The  logarithm  of  1  is  0. 

For  a0  =  1  for  all  values  of  a  ;  therefore  log  1=0,  what- 
ever the  base  may  be. 

(2)  The  logarithm  of  the  base  itself  is  1. 
For  a1  =  a ;  therefore  loga«  =  1. 

(3)  The  logarithm  of  0  in  any  system  whose  base  is  greater 
than  1  is  minus  infinity. 

For  a~°°  =  —  =  —  =  0;  therefore  logO  =  —  a>. 

a00        go 

Also,  the  logarithm  of  -+-  <x>  is  +  co. 

(4)  The  logarithm  of  a  product  is  equal  to  the  s»m  of  the 
logarithms  of  its  factors. 

For  let  mn  be  the  product ;  let  a  be  the  base  of  the  sys- 
tem, and  suppose 

x  =  log?>i,        y  =  logn.  " 

Then  m  =  a1,  n  =  a'J. 

Therefore  the  product,  mn  =  a*  X  a"  =  ax  +  v; 
whence,  by  definition  (Art.  202), 

log?>iH  =  x  +  y  =  logm  +  logx. 
Similarly,  \ogmnp  =  logm  -f  log  n  +  logp; 
and  so  on  for  any  number  of  factors. 
Thus,  log  30  =  log  (2  x  3  X  5) 

=  log  2  +  logo  +  log 5. 


PROPERTIES   OF  LOGARITHMS.  435 

(5)  The  logarithm  of  a  fraction  is  equal  to  the  logarithm 
of  the  numerator  minus  the  logarithm  of  the  denominator. 

For  let  —  be  the  fraction,  and  suppose 
n 

x  =  log?u,         y  =  logn. 
Then  m  =  ax,  n  =  a". 

Therefore  the  fraction,  —  =  —  =  af~y ; 
n       a" 

whence  by  definition, 

log  —  =  x  —  y  =  logm  —  logn. 
n 

Thus,        log  -4/  =  log  42  —  log  5 

=  log  2  -f  log  3  +  log  7  —  log  5. 

(G)  The  logarithm  of  any  power  of  a  number  is  equal  to 
the  logarithm  of  the  number  multiplied  by  the  exponent  of  the 
power. 

For  let  x  =  log  ra  ;  then  m  =  ax. 

Therefore  m?>  =  («x)p  =  apx ; 

whence  by  definition, 

log  mp  =  px  =  p  log  m. 

(7)  The  logarithm  of  any  root  of  a  number  is  equal  to  the 
logarithm  of  the  number  divided  by  the  index  of  the  root. 

For  let  x  =  log  m  ;  then  m  =  ax. 

]  1  X 

Therefore  mr  =  (ax)r  =  ar ; 

whence  by  definition, 

-         x       1 
logfwi'-)  =  -  =  -logm. 
r       r 

It  follows  from  these  results  that  by  the  use  of  logarithms, 
the  operations  of  multiplication  aud  division  may  be  replaced 
by  those  of  addition  and  subtraction  ;  and  the  operations  of 
involution  and  evolution  by  those  ot  multiplication  and 
division. 


436      COMPARISON   OF  TWO  SYSTEMS   OF  LOGARITHMS. 


EXAMPLES. 
1 .    Express  the  logarithm  of  5L_  m  terms  of  log  a,  log  bx 


and  logc. 

loo; 

&  c662 

=  log —  = 
°c562 

log  a* 

-  log(c562) 

=  f  log  a  - 

(logc5 

+  log&2) 

=  flog  a  — 

5  log  c 

-  2  log  6. 

Express  the  following  logarithms  k 

i  terms  of  loga,  logfr, 

and 

logc 
log 

Ans.  G  log  « 

2. 

y/(a2&8)6 

+  9  log  b. 

3. 

log 

(V«'2  x 

\/b*).__ 

flog  a 

+  f  log  &, 

4. 

log- 

{\Ja~2b 

x  Va^-3)- 

—  flog  a 

-  £log&. 

5. 

log 

\Jab~] 

Lc~2 

£  log  a. 

6. 

log 

(V«-v 

&•  +  v'6  V«) 

—  •Hog  a. 

204.   Comparison  of  Two  Systems  of  Logarithms. 

Given  the  logarithm  of  a  number  to  base  a,  to  find  the  loga- 
rithm of  the  same  number  to  base  b. 

Let  m  be  any  number  whose  logarithm  to  base  b  is  required. 

Let  x  =  logbm  ;  then  bx  =  m ; 

.-.     \oga(bx)  =  \ogam  ;  ora;log0&  =  loga?u. 

•  •.     x  = x  logam, 

loga6 

,  log-m  , , , 

or  Io<j-,,w  =      &2— 0) 

logao 

Hence,  to  transform  the  logarithm  of  a  number  from  base 

a  to  base  b,  tve  mult i '.ply  il  by . 

loga6 

Putting  a  for  m  in  (1).  we  have 

lo&a  =  |2&|  =      *       by  (2)  of  Art.  203. 
.-.    loeba  x  loff.,6  =  1. 


COMMON  SYSTEM   OF  LOGARITHMS.  437 

205.  Common  System  of  Logarithms.  —  Although 
there  may  be  any  number  of  systems  of  logarithms,  there 
are  only  two  in  general  use,  viz.,  the  Naperian*  system  and 
the  common  system. 

The  Naperian  system  is  used  for  purely  Algebraic  pur- 
poses;  its  base  is  2.7182818. 

The  common  system  of  logarithms  is  the  only  system  that 
is  used  in  practical  calculations;!  its  base  is  10.  Hence, 
since 


10°  =  1, 

log  1  =  0. 

101  =  10, 

log  10  =  1. 

102  =  100, 

log  100  =  2. 

103  =  1000, 

log  1000  =  3. 

104  =  10000, 

log  10000  =  4. 

io-1  =    Tv    = 

•1, 

log  .1  = 

-1. 

10  -2=       yh       = 

.01, 

log  .01  = 

-2. 

10  "3=    toW    = 

.001, 

log  .001  = 

-3. 

1A-4    1            

1U         —   Too  (TIT   — 

.0001, 

log  .0001  = 

-4. 

From  an  inspection  of  these  examples  it  is  evident  that, 
in  the  common  system,  the  logarithm  of  any  number  between 

1  and  10  is  some  number  between  0  and  1  ;  i.e.j  0  -f-  a 
fraction, 

10  and  100  is  some  number  between  1  and  2  ;  i.e.,  1  -f  a 
fraction, 

100  and  1000  is  some  number  between  2  and  3  ;  i.e.,  2  +  a 
fraction, 

1  and  .1  is  some  number  between  0  and  —1  ;  i.e.,  —1  -f-  a 
fraction, 

.1   and    .01    is   some  number  between  —1  and  —2;  i.e., 

—  2  +  a  fraction, 

.01  and  .001  is  some  number  between  —2  and  —3;  i.e., 

—  3  +  a  fraction. 


*  So  called  from  its  inventor,  Baron  Napier,  a  Scotch  mathematician. 
t  This  system   was   first   introduced,   iu  1615,    by   Briyya,  a  contemporary   ol 
Napier. 


438  COMMON   SYSTEM   OF  LOGARITHMS. 

It  thus  appears  that  in  general  the  common  logarithm  of  a 
number  consists  of  two  parts,  an  integral  part  and  a  decimal 
part.  The  integral  part  is  called  the  characteristic,  and  the 
decimal  part  is  called  the  mantissa. 

From  the  above  examples  we  see  that  if  a  whole  number 
is  expressed  by  one  digit,  the  characteristic  of  its  logarithm 
is  0  ;  if  it  is  expressed  by  two  digits,  the  characteristic  of  its 
logarithm  is  1  ;  if  it  is  expressed  by  three  digits,  the  char- 
acteristic is  2  ;  and  so  on,  the  characteristic  being  always  one 
less  than  the  number  of  digits  in  the  number.  Also  the 
logarithm  of  any  number  between  1  and  .1  has  —1  for  its 
characteristic;  the  logarithm  of  an}'  number  between  .1  and 
.01  has  —2  for  its  characteristic ;  the  logarithm  of  any  num- 
ber between  .01  and  .001  has  —8  for  its  characteristic;  and 
so  on,  the  characteristic  being  always  negative  and  one 
greater  than  the  number  of  ciphers  immediately  after  the 
decimal  point. 

In  general.  Let  N  be  a  number  whose  integral  part 
contains  n  digits  ;  then 

jy    _    JQ(n-l) +afraction_ 

.*.     log  A7  =  (n  —  1)  +  a  fraction. 
Also,  let  D  be  a  decimal  with  n  ciphers  immediately  after 
the  decimal  point ;  then 

D  =  io -("hu +*«>". 
logZ)  =  —  (n  +  1)  +  a  fraction. 
Therefore  the  characteristic  of  the  common  logarithm  of 
any  number  may  be  written  down  by  the  following 
Rule.  % 

(1)  Tlie  characteristic  of  the  logarithm  of  a  number  greatei 
than  unity  is  posit  ire,  and  less  by  one  than  the  number  of 
digits  in  its  integral  part. 

(2)  The  characteristic  of  the  logarithm  of  a  decimal  /rue/inn 
is  negative,  and  greater  by  one  than  tin'  number  of  ciphers 
immediately  after  the  decimal  point. 

Thus,  the  characteristic  of  log  2347.3  is  8  j  characteristic 
of  log .000459  is  -4. 


TABLE   OF  LOGARITHMS.  439 

206.  Table  of  Logarithms.  —  The  common  logarithms 
of  all  integers  from  1  to  200000  have  been  found  and 
tabulated.  In  most  tables  they  are  given  to  six  places  of 
decimals  ;  and  tables  of  logarithms  to  seven  places  of  deci- 
mals are  in  common  use  for  Astronomical  and  Mathematical 
calculations.  This  is  the  system  in  practical  use,  and  it  luis 
two  great  advantages : 

(1)  From  the  rule  (Art.  205)  the  characteristics  can  be 
written  down  at  once,  so  that  only  the  mantissa}  have  to 
be  given  in  the  tables. 

(2)  The  mantissa?  are  the  same  for  the  logarithms  of  all 
numbers  which  have  the  same  significant  digits  ;  so  that  it 
is  sufficient  to  tabulate  the  mantissse  of  the  logarithms  of 
integers. 

For,  since  altering  the  position  of  the  decimal  point 
without  changing  the  sequence  of  figures  merely  multiplies 
or  divides  the  number  by  an  integral  power  of  10,  it  follows 
that  its  logarithm  will  be  increased  or  diminished  by  an 
integer,  i.e.,  that  the  mantissa  of  the  logarithm  remains 
unaltered. 

In  general.  If  N  be  any  number,  and  p  and  q  any  in- 
tegers, it  follows  that  N  X  10p  and  N  -=-  1017  are  numbers 
whose  significant  digits  are  the  same  as  those  of  N. 

Then  log  (N  x  10")  =  \ogN  +  p  log  10  =  log  N  +  p  .   (1 ) 

Also  log  (N  +  10?)  =  log  A7"  -  q  log  10  =  log  N  -  q  .   (2) 

In  (1)  the  logarithm  of  A7"  is  increased  by  an  integer,  and 

in  (2)  it  is  diminished  by  an  integer ;   that  is,  the  mantissa, 

or  decimal  portion  of  the  logarithm,  remains  unchanged,  a 

property  which  is  true  only  in  the  common  system. 

Thus,  from  a  table  of  logarithms  we  find  the  mantissa  of 

the  logarithm  of  2583  to  be  412124  ;  therefore,  prefixing  the 

characteristic  with  its  appropriate  sign  according  to  the  rule, 

we  have  log  2583  =  3.412124, 

log  258.3  =  2.412124, 

log  25.83  =  1.412124, 

log  2.583  =  0.412124. 


440  EXAMPLES. 

The  mantissa  is  always  kept  positive.  In  the  case  of  a 
negative  logarithm  the  minus  sign  is  written  over  the  charac- 
teristic, and  not  before  it,  to  indicate  that  the  characteristic 
alone  is  negative,  the  mantissa  being  always  positive. 

Thus,  3. 30103,  the  logarithm  of  .002,  is  equivalent  to 
—  3  +  .30103,  and  must  be  distinguished  from  —3.30103,  an 
expression  in  which  both  the  integer  and  the  decimal  are 
negative.  Sometimes  in  working  with  negative  logarithms 
an  Arithmetic  artifice  will  be  necessary  to  make  the  mantissa 
positive.  For  example,  a  result  such  as  —2.69897,  in  which 
the  whole  expression  is  negative,  may  be  transformed  by 
subtracting  1  from  the  characteristic,  and  adding  1  to  the 
mantissa.     Thus, 

-2.69897  =  -3  +  (1  -  .69897)  =  3.30103. 

Note.  —  When  the  characteristic  of  a  logarithm  is  negative,  it  is 
often,  for  convenience,  made  positive  by  the  addition  of  10;  which 
can  lead  to  no  error,  if  we  are  careful  to  subtract  10. 

Thus,  instead  of  the  logarithms  3.603582  and  4.026011,  we  may 
write  7.6035S2  —  10  and  6.026011  -  10. 


EXAMPLES. 

The  following  examples  will  illustrate  the  utility  of  loga- 
rithms in  facilitating  Arithmetic  calculations.  For  infor- 
mation as  to  the  use  of  Logarithmic  Tables  the  student  is 
referred  to  works  on  Trigonometry  and  to  the  introductions 
to  Tables  of  Logarithms. 

1.  Required  the  logarithm  of  .0002432. 

In  the  tables  of  logarithms  we  find  that  3859636  is  the 
mantissa  of  log  2432  (the  decimal  point,  as  well  as  the  char- 
acteristic, being  omitted);  and  by  the  rule  (Art.  205).  the 
characteristic  of  the  logarithm  of  the  given  number  is  —4. 

.-.     log  .0002432  =  4.3859636. 

2.  Find  the  value  of  \/.00000165,  given 

log  165  =  2.2174889,     ami     log  697424  =  5.8434968. 


EXAMPLES.  441 

Let  x  =  the  value  required  ;  then 
log  a-  =  log  (.00000165)*  =  \  log  (.00000165)  [Art.  203,  (7)] 
=  1(6.2174839)        =  £(10  +  4.2174839) 
=  2.8434968, 
and  .8434968  is  the  mantissa  of  log  697424;    hence  k  is  a 
number  consisting  of  these  same  digits,  but  with  one  cipher 
after  the  decimal  point,  by  the  rule  (Art.  205). 
.-.     x  =  .0697424. 
It  will  be  seen  that  we  subtracted  4  from  the  characteristic 
—  6  to  make  it  exactly  divisible  by  the  index  5,  and  there- 
fore we  added  the  same  number  4  to  the  mantissa,  to  keep 
the  value  of  the  logarithm  unchanged. 

3.  Given  log  3  =  . 4771213;  findlog  [(2.7)8x(.81)*-=-(90)l]. 
The  required  value  =  3  logfj  +  f  ^StVo  ~~  I  log  90 

=  3(log33  -  1)  +  f(log34  -  2)  -  £(log32  +  1) 
=  (9  +  ¥  -f)log3  -(3  +  f  +  |) 
=  ?■£  log  3  -  5i£  =  4.6280766  -  5.85 
=  2.7780766. 

Note.  —  The  logarithm  of  5  and  its  powers  can  always  be  obtained 
from  log  2 ;  thus 

log5  =  log^  =  log  10  —  log2  =  1  -  log2. 

4.  Find  the  number  of  digits  in  87516,  given 

log  2  =  .3010300,  and  log  7  =  .8450980. 
log(87516)  =  16  log  (7  x  125)  =  16(log7  +  3  log  5) 
=  16(log7  +  3-3  log 2)  =  47.072128; 
hence  the  number  of  digits  is  48  (Art.  205). 

Given  log  2  =  .3010300, log 3  =.4771213, log  7  =.8450980  ; 
find  the  value  of 


8.  log  4f.     Ans.  .6690067. 

9.  log V 12.  .3597271. 
10.  log  V- 01 05.     1.5052973. 

11.  Given  log  2  and  log  3  ;  find  Log  V/ ~.       1.1998692. 


5.  log 64.  Ans.  1.8061800. 

6.  log.  128.   1.1072100. 

7.  W.0125.  2.0909100. 


442  EXPONENTIAL   EQUATIONS  —  EXAMPLES. 

207.  Exponential  Equations.  —  An  Exponential  Equo 
Hon  is  one  in  which  the  exponent  is  the  unknown  quantity. 
Such  equations  are  easily  solved  by  logarithms.     Thus, 

1.  Given  ax  =  b  ;  find  the  value  of  x. 

Taking  the  logarithms  of  both  members,  we  have 
x  log  a  =  log  b 
log& 

X  =   — 2 — . 

log  a 

2.  Find  the  value  of  x  in  2X  =  128. 
Here  a;  log  2  =  log  128  =  log27  =  7  log  2. 

"     X~    log2    ~7' 

3.  Find  the  logarithm  of  32y/4  to  base  2y/2. 

Let  x  be  the  required  logarithm  ;  then  by  definition  (Art. 
202),  (2^2)*  =  32y/4  =  25  ■  2*. 

Taking  logarithms  of  both  sides,  we  have 

x log  2*  =  log  25° ,  or  fa:  log  2  =  %7  log  2. 

•••    «  =  ¥  +  f  =  ¥  =  3.6. 

4.  Given  log  2  and  log  3  (Art.  206,  Ex.  5)  ;  find  to  two 
places  of  decimals  the  value  of  x  from  the  equation 

Taking  logarithms  of  both  sides,  we  have 

(3  -  4a;)  log  G  +  (a;  +  5)  log  4  =  log  8  ; 
.-.     (3  -  Ax)  (log  2  +  log  3)  +  (a  +  5)2  log  2  =  3  log  2; 
.-.    x{—  4  log2  -4 log 3  +  21og2)=:  ?,  log2  -  31og2  -  3  log3  -  101og2; 
x  =  10  log  2  +  3  log  3  =  4.4410639  _  j  7? 
2  log  2  -t-  41og3       2.5105452 

Solve  the  following  equations : 

5.  ax  =  cbx.  Ans. 


a  -  los& 


EXPONENTIAL   SERIES. 

6.  a2^3*  =  c6.  Ans. 

7.  a' 


443 


5  lop;  c 


2  log  a  +  3  log  6 

)og(logc)  —  log  (log  a) 

log  b 

bx 
Note.  —  In  this  example,  by  a     is  meant  a  raised  to  the  power 


c. 


expressed  by  bx,  and  not  ab  raised  to  the  Xth  power. 


8.    C" 


<tb" 


Ans. 


9.    a2*  -  2ax 


8. 


log  a  —  log  6 

2 2 . 

Ml  log  C  —  11  log  6 

2  log  2 
loga  ' 
Find  the  logarithms  of 

10.  16  to  base  ^2,  and  1728  to  base  2V^3  Ans.  8,  6. 

11.  125  to  base  5^5,  and  .25  to  base  4.  2,  —1. 

12.  ^|¥  to  base  2\/2,  and   £  to  base  9.  —  *-£,  -| 

Find  the  value  of 

13.  log8128,  andlog6^.  Ans.  2$,  -3. 

Solve  the  following  equations,  having  given  log  2,  log  3, 
log  7  (Art,  20G,  Kx.  5): 


14.  3* 

15.  5* 


203. 


-4ns.  3.46. 
4.29. 


16.  55-s*  =  2*  +  2.  4ns.  1.206. 

17.  2F=22r  +  1.5*.      14.206. 


EXPONENTIAL    AND    LOGARITHMIC    SERIES. 

208.   Exponential  Series.  —  To  expand  ax  in  a  series 
of  ascending  powers  of  x. 

By  the  Binomial  Theorem  we  have 


w.t-2)  1    , 
— i  +  •  •  • 


A     l-V*     i.       1    nx(nx-l)   1  ,  nx(nx—l)(r, 

(1+-)   =l+n»-+— ^— i-i+— i ~^ 

\      nj  n  [2_        ?i2  [_3_ 

,   *H)  -KX-3 

=  l+a?+.  V       7+— -^ -+ 


(1) 


444  LOGARITHMIC  SERIES. 

(1+«T=[(l+!)7=[1+1+Tf+L4^+-P 

and  therefore  series  (1)  is  equal  to  series  (2)  however  great 
n  may  be.  Hence  if  n  be  indefinitely  increased,  we  have 
from  (1)  and  (2) 

l+x+l?L+£?.+^L  +  .  .  .  =  /l  +  l-fi_  +  i_+J_+  .  .    V 

LI    ^.    Li  V         LA    LI    Li         y 

The  series  in  the  parenthesis  is  usually  denoted  by  e ; 
■-™,..1+. +  £+£  +  £  +  ..  ..,      .     .    (3) 

Li     Ll     Li 

which  is  the  expansion  of  ex  in  powers  of  a;;  to  find  the 
expansion  of  ax,  let  a  =  ee,  so  that  c  =  loge  a.  Substituting 
in  (3)  we  have 

,,20.2  „S.-,S  ~40.4 

ax  =  ecx  =  1  +  ex  +  —  +  —  +  —  + 

LA      UL      Li 

or 
0.=1+(lo&a)*+i^^+i!^^:+fl^>^+...  (4, 

This  result  is  called  the  Exponential  Theorem. 
If  we  put  x  =  1,  we  have  from  (3) 

+  L1     LI     Li     LI     L^L 

From  this  series  we  may  readily  compute  the  approximate 
value  of  e  to  any  required  degree  of  accuracy.  This  con- 
stant value  of  e  is  called  the  Naperian  base  (Art.  205).  To 
ten  places  of  decimals  it  is  found  to  be  2.718281.8284. 

209.  Logarithmic  Series.  —  To  expand  %e(l  +  a;)  in 
a  series  of  ascending  powers  of  x. 

By  the  Binomial  Theorem  we  have 
a*  =  (1  +  o  -  1)*  =  1  +  x{a  -  1)  +  X(X~X\a-  -  \y 

+  <—iK«-i)(g_1)1  +  _ 

Bl+a;[a-l-|(o-i)»+Ko-l)*-Ka- !)*+•••] 

+  terms  involving  x~,  sc",  etc. 


COMPUTATION  OF  LOGARITHMS.  445 

Comparing   this  value  of   ax  with  that   given   in   (4)   of 
Art.  208,  and  equating  the  coefficients  of  x,  we  have 
logea  =  a  -  1  -  |(a  -  1)'J  +  £(«  ~  1)3  -  £(«  -  l)4  +  •  •  • 

Put  a  =  1  ■+-  x ;  then 

loge(l  +  a)  =  *  -  |2  +  |3  -  -|  + (1) 

This  is  the  Logarithmic  Series;  but  unless  x  be  very  small, 
the  terms  diminish  so  slowly  that  a  large  number  of  them 
will  have  to  be  taken  ;  and  hence  the  series  is  of  very  little 
use  for  numerical  calculations.  If  x  =  1,  the  series  is 
altogether  unsuitable.  We  may,  however,  deduce  from  it 
other  series  by  the  aid  of  which  Tables  of  Logarithms  may 
be  constructed. 

Changing  x  into  — jc,  (1)  becomes 

loge(l  -  x)  =  -*  -  |V  |  -  t  - (2) 

Subtracting  (2)  from  (1),  we  have 

loge(l  +  c)  -  logc(l  -  x)  =  2x  +  y +~  + 


,       1  4-  x       J     .   a;3   .    x5   .   x1    ,  \ 


(3) 


Put  LilJL;  =  ?L±i,  and       .-.    aj  = ;    (3)   thus 

1  ~  x  n  2n  -f-  1      v  ' 

becomes 

gfi     w        "    [_2w  +  l       3(2%  +  l)3       5(2w  +  l)s  J 

or 

1<)g.(B+l)  =  lC)g.n+2[^+3li^+675iI?+...](4) 

210.  Computation  of  Logarithms.  —  By  means  of 
series  (4)  of  Art.  209,  the  logarithm  of  either  of  two  con- 
secutive numbers  may  be  computed  to  any  extent  when  the 
logarithm  of  the  other  number  is  known,  and  thus  a  table 
of  logarithms  can  be  constructed.     Logarithms  to  the  base  e 


446  NAPERIAN  LOGARITHMS. 

are  called  Naperian  Logarithms  (Art.  205).  They  are  also 
called  natural  logarithms,  because  they  are  the  first  log- 
arithms which  occur  in  the  investigation  of  a  method  of 
calculating  logarithms.  When  logarithms  are  used  in  theo- 
retical investigations,  the  base  e  is  always  understood,  just 
as  in  practical  calculations  the  base  10  is  invariably  employed. 
It  is  only  necessary  to  compute  the  logarithms  of  pHrru 
numbers  from  the  series,  since  the  logarithm  of  a  composite 
number  may  be  obtained  by  adding  together  the  logarithms 
of  its  component  factors.  The  logarithm  of  1  is  0.  Making 
u  =  1,  2,  4,  6,  etc.  successively  iu  (4)  of  Art.  209,  we 
obtain  the  followiug 

NAPERIAN     LOGARITHMS. 

log,  2  =  log,  1  +  2  (-  +  —  +  —  -  +  -L-  +  —  +  .  .  .  V 

or,  since  loge  1  =  0, 

loge2  =  2(.33333333  +  .01234568  +  .00082305  +  .00006532 

+  . 00000565 +. 00000051  -f  .00000005+.  .  .) 
=±2(0.34657359)  =0.69314718. 

loge3==loge2  +  2A+^i-4--Vr  +  ^i-+.  .  \  =  1.09861228. 

\.>     3  •  o3     5  •  5d     7  •  o7  / 

loge4  =  21oge2  =1.38029436. 

logc5  =  loge4  +  2/'-+— !-H — L .-J — L_+.    A= 1.60943790. 
oe    T  \9    8-9*    5-95    7-97  / 

loge  6  =  log,  3+loge  2  =  1 .  79 1 75946. 

loge7  =  loge6  +  2^--  +  -1—+-1—  +  .  .  .\        =1.94590996. 

loge8  =  31oge2  =2.07944154. 

logc9  =  21og,,3  =2.19722456. 

logc  I0=loge5+logc2  =2.80258509. 

In  this  way  the  Naperian  logarithms  of  all  numbers  may 
be  computed.    Where  llnv  numbers  are  large,  their  logarithms 

are  computed  mure  easily  thau  in  the  case  of  small  numbers. 


INTEREST  AND   ANNUITIES  —  SIMPLE   INTEREST.    447 

Thus,  in  computing  the  logarithm  of  101,  the  first  term  of 
the  series  gives  the  result  true  to  seven  places  of  decimals. 
By  changing  b  to  10  and  a  to.e  in  (1)  of  Art.  204,  we  have 

logMm  =  kg^L  =       log.™       =  .43429448  log.m, 
010  ]oge10        2.30258509  oe     ' 

or         common  log  m  =  Naperian  log  m  x  .43429448. 

The  number  .43429448  is  called  the  modulus  of  the  common 
system. 

Hence,  the  common  logarithm  of  any  number  is  equal  to 
the  Naperian  logarithm  of  the  same  number  multiplied  by  the 
modulus  of  the  common  system,  .43429448. 

Conversely.  Having  given  the  common  logarithm  of  a 
number,  to  find  its  Naperian  logarithm;  divide  the  common 
logarithm  by  .43429448. 

Note.  —  The  base  e,  the  modulus  of  the  common  system,  and  the 
Naperian  logarithms  of  2,  3,  and  5  have  all  been  calculated  to  more 
than  260  places  of  decimals.  See  the  Proceedings  of  the  Royal  Society 
of  London,  Vol.  XXVII,  p.  88. 

Denoting  the  modulus  by  M,  and  multiplying  (4)  of  Art. 
209  by  it,  we  obtain  a  series  by  which  common  logarithms 
may  be  computed  ;  thus, 

1.  Find  by  means  of  (1)  the  common  logarithms  of  65 
and  131  ;  given  log  2  =  .3010300,  and  M  =  .43429448. 

Ans.  1.8129134,  2.1172713. 

2.  Find  the  Naperian  logarithm  of  1325.07. 

Ans.  7.18923. 

INTEREST    AND    ANNUITIES. 

211.  Simple  Interest.  —  To  find  the  interest  and  amount 
of  a  given  sum  in  a  given  time  at  simple  interest. 

Interest  is  money  paid  for  the  use  of  money.  The  Princi- 
pal is  the  sum  lent.  The  Amount  is  the  sum  of  the  Principal 
and  Interest  at  the  end  of  the  time.  When  interest  is  paid 
on  the  principal  alone,  it  is  called  Simple  Interest. 


448  PRESENT    VALUE  AND   DISCOUNT. 

Let  P  be  the  principal  in  dollars,  r  the  interest  of  $1  for 
one  year,  n  the  number  of  years,  I  the  interest,  and  M  the 
amount. 

Since  the  interest  of  %\  for  one  year  is  r,  the  interest  of 
P  for  one  year  is  Pr,  and  therefore  for  n  years  it  is  Pur; 
that  is. 

I  =  Pnr (1) 

And  for  the  amount,  M  =  P  +  I,  or  M  =  P(l  +  nr),  (2) 

From  (1)  and  (2),  if  any  three  of  the  quantities  P,  n,  r,  7, 
or  P,  n,  r,  M,  are  given,  the  fourth  may  be  found. 

212.  Present  Value  and  Discount.  —  To  find  the 
present  value  and  discount  of  a  given  sum  due  at  the  end  of 
a  given  time,  at  simple  interest. 

The  Present  Value  of  a  sum  due  at  the  end  of  a  given 
time,  is  the  principal  which  with  its  interest  for  the  given  time 
will  amount  to  the  sum. 

The  Discount  is  the  difference  between  the  sum  and  its 
present  value. 

Let  M  be  the  given  sum  or  amount,  r  the  interest  of  $1 
for  one  year,  n  the  number  of  years,  P  the  present  value, 
and  D  the  discount. 

.Since  P  is  the  sum  which  will  in  n  years  amount  to  il/,  we 
have  from  (2)  of  Art.  211, 

M  =  P(l  +  nr)  ;         .-.     P  =       M       .     .    (1) 

1  -f-  nr 

And  Z>=  M-P=     Mnr      .     .   (2) 

1   +  nr 

Note. — This  value  of  7)  is  called  the  true  discount.  In  practice, 
when  a  sum  of  money  is  paid  before  it  is  due,  it  is  customary  to  deduct 
the  interest  on  the  debt  instead  of  the  true  discount,  and  the  money 
so  deducted  is  called  the  banker's  discount;  tlms, 

Banker's  Discount  =  Mnr.     True  Discount  = — .  (3) 

1  +  nr  K  } 

1.    Required  the  rate  when  $300  in  2   years   at   simple 

interest  amounts  to  $318. 


COMPOUND   INTEREST.  449 

Here  M  -  318,  P  =  300,  n  =  2.  Solving  (2)  of  Art. 
211  for  r  and  substituting,  we  have 

318  -  300         nQ  .  .m      0 

r  =  =  .03.         .•.     rate  per  cent*  =3. 

300  x  2 

2.  The  difference  between  the  true  discount  and  the  bank 
discount  on  $1900  paid  4  months  before  it  is  due  is  33^  cents  : 
find  the  rate  per  cent. 

Here  M  =  $1900,  and  n  =  *;  therefore  from  (3)  we 
have 

1900r 


1900r 


h 


3  1  + 

Solving,  we  get  r  =  .04.         .•.     rate  per  cent  =  4. 

213,  Compound  Interest.  —  To  find  the  interest  and 
Amount  of  a  given  sum  in  a  given  time  at  compound  interest. 

When  interest  as  soon  as  it  becomes  due  is  added  to  the 
principal,  and  interest  charged  upon  the  whole,  it  is  called 
Compound  Interest. 

Let  P  be  the  principal,  11  the  amount  of  $1  in  one  year, 
so  that  II  =  1  +  r,  n  the  number  of  years,  /  the  interest, 
and  M  t4ie  amount. 

Since  the  amount  of  $1  for  one  year  is  E,  the  amount 
of  P  at  the  end  of  the  first  year  is  Pli  ;  and  since  this  is 
the  principal  for  the  second  year,  the  amount  at  the  end  of  the 
second  year  is  PR  x  R  or  PR2.  Similarly  the  amount  of 
PR2  at  the  end  of  the  third  year  is  PRS,  and  so  on  ;  hence 
the  amount  of  P  in  n  years  is  PR" ;  that  is, 

M=PR\         .-.     I  =  P(R"  -  1)  .     .     .   (1) 

If   the  interest  is  pa}*able  more  frequently  than   once  a 

*  It  must  be  borne  iu  mind  that  r  is  not  the  rate  per  cent,  but  only  the  hundredth 
part  of  it. 


450  PRESENT    VALUE  AND   DISCOUNT. 

year ;    for  example,    if   it   is    payable    semi-annually,   the 
interest  for  $1  for  half  a  year  will  be  -. 

.-.     The  amount  of  Pin  half  a  year  =  P(l  +  -)• 

The  amount  of  P  in  one  year  =P\  1  +  -  )  • 

The  amount  of  P  in  n  years  =P[  1  +  -)   • 
Similarly  if  the  interest  is  payable  quarterly. 
The  amount  of  P  in  n  years  =P(l+-\  . 

And  generally,  if  the  interest  is  payable  q  times  a  year  at 
equal  intervals,  the  interest  of  $1  for  each  interval  will  be  -. 

1-1 — )  • 

If  the  interest  is  due,  i.e.,  converted  into  principal,  every 
moment,  then  q  becomes  infinitely  great,  i.e.,  the  intervals 
between  the  payments  are  infinitely  small.    To  find  the  value 

of  the  amount,  put  -  =  -,  so  that  q  =  rx ;  thus  the  amount 
q       x 

of  P  in  n  years 

-^7-^-*+ ST 

=  Pe"r  [(2)  and  (3)  of  Art.  208], 
since  x  is  infinite  when  q  is  infinite. 

214.    Present   Value    and    Discount. —  To  find   the 

present  value  and  discount  of  a  given  sum  due  at  the  end  of  a 
given  time,  at  compound  interest. 

Let  M  be  the  given  sum  or  amount,  P  the  present  value. 
7)  the  discount,  A'  the  amount  of  §1  for  one  year,  n  the 
number  of  years. 


PRESENT    VALUE  AND   DISCOUNT.  451 

Since  P  is  the  sum  which  in  n  years  will  amount  to  31,  we 
have  (Art.  213) 

M  =  PR 

...     P=K.         D=M-P=M^n-'\    .   (1) 

1.  Find  the  amount  of  $100  in  one  year  at  5  percent, 
when  the  interest  is  converted  into  principal  semi-annually. 

Here  P  =  100,  /•  =  .05,  n  =  1,  and  1  +  -  =  1.025. 
.-.     Amount  =  100(1.025)2  =  $105.0025. 

2.  The  present  value  of  $G72  due  in  a  certain  time  is  $126  ; 
if  compound  interest  at  4£  per  cent  be  allowed,  find  the  time, 
having  given  log  2  =  .30103,  and  log  3  —  .47712. 

Here       r  =  .04  J  =  fo  and  R  =  1  -f  r  ==  §£. 

Let  n  =  the  number  of  years  ;  then  by  (1)  of  Art.  213, 
we  have 

672  =  126(|5)«;         ...     n  log f|  =  log1^. 

.-.     n  (log  100  -  log  DO)  =  log  16  -  log  3. 

4  lose  2  -  log3  .72700        „  . 

•  •     n  ~  \ ^ ^ > :  —    /^„F.,  =  41,  very  nearly; 

2  -  5  log  2  -  log  3        .01773  J  J 

thus  the  time  is  very  nearly  41  years. 

3.  At  simple  interest,  the  interest  on  a  certain  sum  of 
money  is  $180,  and  the  discount  on  the  same  sum  for  the 
same  time  and  at  the  same  rate  is  $150  :  find  the  sum. 

Ans.  $900. 

4.  If  a  sum  of  money  doubles  itself  in  40  years  at  simple 
interest,  find  the  rate  of  interest.  Ans.  2|. 

5.  Find  in  how  many  years  $100  will  become  $1050  at  5 
per  cent  compound  interest ;  having  given  log  1 1  =  1.14613, 
log  15  =  1. 17G0D,  log  16  =  1.20412. 

Ans.  Between  48  and  li). 


452  Till:   AMOUNT   OF  AN  ANNUITY. 

215.  Annuities.  —  An  annuity  is  a  fixed  sum  payable 
at  equal  intervals  of  time  under  certain  stated  conditions. 

An  annuity  certain  is  one  which  is  payable  for  a  fixed 
number  of  years  independent  of  any  contingency.  A  life 
annuity  is  one  which  is  payable  during  the  lifetime  of  a 
person,  or  of  the  survivor  of  a  number  of  persons.  A 
deferred  annuity,  or  reversion,  is  an  annuity  which  does  not 
begin  until  after  the  lapse  of  a  certain  number  of  years  ;  and 
when  the  annuity  is  deferred  for  p  years,  it  is  said  to  com- 
mence after  p  years,  and  the  first  payment  is  made  at  the 
end  of  p  +  1  years. 

A  perpetuity  is  one  which  is  to  continue  forever ;  if  it  does 
not  commence  at  once  it  is  called  a  deferred  perpetuity.  An 
annuity  left  unpaid  for  a  certain  number  of  years  is  said  to 
be  forborne  for  that  number  of  years. 

216.  The  Amount  of  an  Annuity.  —  To  find  the 
xmount  of  an  annuity  left  unpaid  for  a  given  number  of 
years. 

( 1 )  At  simple  interest. 

Let  ^4  be  the  annuity,  n  the  number  of  years,  r  the  interest 
of  $1  for  one  year,  M  the  amount. 

At  the  end  of  the  first  year  A  becomes  due,  and  the 
amount  of  this  sum  in  the  remaining  n  —  1  years  is  A  + 
(n  —  \)Ar ;  at  the  end  of  the  second  year  another  A  is  due, 
and  the  amount  of  this  sum  in  the  remaining  n  —  2  years 
is  A  +  (n  —  2)rA;  and  so  on  to  the  end  of  the  ?i,h  year. 
Hence,  calling  M  the  sum  of  all  these  amounts,  we  have 

M  =  A+(n-l)rA+A+(n-2)rA+ +A+rA+A 

=  nA  +  (1  +  2  +  3  + +  n  -  l)rA 

=  nA  +  w(M  ~  })  rA.     [(:;)  of  Art.  L61]    ...   (1) 

(2)  At  compound  interest. 

Let  11  be  the  amount  of  $1  for  one  year.  At  the  end  of 
the  firsl  year  .1  is  due  and  the  amount  of  this  sum  in  the 
remainingn  —  1  years  is  AU""i  (Art.  218)  ;  at  the  end  of 


THE  PRESENT    VALVE   OF  AN  ANNUITY.  453 

the  second  year  another  A  is  clue,  and  the  amount  of  this  sum 
in  the  remaining  n  —  2  years  is  AR"~2 ;  and  so  on  to  the 
last  annuity  payable  at  the  end  of  the  uth  year.  Hence 
the  entire  amount  due  at  the  end  of  n  years  is  the  sum  of 
these  amounts. 
•.  M  =  AH"-1  +  AR'-2  +  ....  +  AR2  +  AR  +  A 
=  (1  +  R  -f-  B2  -f  .  .  .  .  to  n  terms)J. 

=  f-^-y-i.     [(2)  of  Art.  163]       ....    (2) 

217.  The  Present  Value  of  an  Annuity.  —  To  find 
the  present  value  of  an  annuity  to  continue  for  a  given  number 
of  years,  at  compound  interest. 

Note.  —  In  finding  the  present  value  of  annuities,  it  is  always 
customary  to  reckon  compound  interest. 

Let  P  be  the  required  present  value,  A  the  annuity,  R  the 
amount  of  $1  in  one  year,  n  the  number  of  years.  Then 
the  amount  of  P  in  n  years  must  equal  the  amount  of  the 
annuity  in  the  same  time  ;  that  is  (Arts.  213,  210), 

J?'1  —  1  1    —   7?-" 

PRn    =    K  lA.  .-.        P=    l  ^       A.         .      (1) 

R  -  1  R  -  1  v  ' 

If  we  make  n  infinite,  the  annuity  becomes  a  perpetuity 
(Art.  215).  Hence,  making  n  infinite  in  (1),  we  have  for 
the  present  value  of  a  perpetuity, 

A  A 


R  -  1        r 


C-?) 


Rem.  —  If  the  present  value  of  an  annuity  A  for  any  number  of 
years  be  mA,  the  auuuity  is  said  to  be  worth  m  gears'  purchase. 

Iu  the  case  of  a  perpetuity,  mA  =  — ,  by  (2). 

1  ioo 

.-.     m  =  -  = ;  .     .     .     .   (3) 

r       rate  per  cent 

that  is,  the  number  of  years'  purchase  of  a  perpetual  annu- 
ity is  obtained  by  dividing  100  by  the  rate  per  cent. 


454     THE   PRESENT    VALUE    OF  A    DEFERRED    ANNUITY. 

218.  The  Present  Value  of  a  Deferred  Annuity.  — 

To  find  the  'present  value  of  a  deferred  annuity,  to  commence 
at  the  end  of  p  years  and  to  continue  for  n  years,  at  com- 
pound interest. 

Let  A  be  the  annuity,  R  the  amount  of  $1  in  one  year,  P 
the  present  value.  The  present  value  of  an  annuity  to  com- 
mence at  the  end  of  p  years,  and  then  to  continue  n  years, 
is  found  by  subtracting  the  present  value  of  the  annuity  for 
p  years  from  the  present  value  of  the  annuity  for  p  +  n 
years. 

By  (1)  of  Art.  217,  we  have 

i  j?-  p 

the  present  value  of  the  annuity  for  p  years     = A  ; 

the  present  value  of  the  annuity  for  p  +  n  years  = A. 

R — 1 

.-.     P  = ^ A  - ^—  A 

R-\  R  -  1 

AR-p       AR~»-' 


R  -  1         R  -  1 


(1) 


The  present  value  of  a  deferred  perpetuity  (Art.  215)  to 
commence  after  p  years  is 

P=^Z1 (2) 

R  -  1  K  } 

since  n  becomes  infinite. 

A  F 'reeh ■old  Estate  is  an  estate  which  yields  a  perpetual 
annuity  called  the  rent;  and  thus  the  value  of  the  estate  is 
equal  to  the  present  value  of  a  perpetuity  equal  to  the  rent. 

If,  therefore,  we  know  the  number  of  years'  purchase  that 
a  tenant  pays  to  buy  his  farm,  we  obtain  the  rate  per  cent  at 
which  interest  is  reckoned  by  dividing  100  by  the  number  of 
years'  purchase  [Art.  217,  (•'>)]• 

Given  the  following  logarithms  for  use  when  required  : 
log  1.04  =  .017033;    log  1.05  =  .021 189;  log  1.8009  =  .255495; 
log2.19U  =  .340660;  logl.980  =  .296646;  log2.7859  =  .444969 


THE  PRESENT    VALUE   OE  A    DEFERRED   ANNUITY.    455 

1.    An  annuity  of  $500  was  unpaid  for  15  years.     What 
was  the  amount  due,  at  4  per  cent  compound  interest? 

By  (2)  of  Art.  216,  M  =  (^Q4)15  ~  1  x  500  =  $9983. 
J  v  '  .04 


2.  Find  the  present  value  of  an  annuity  of  $500  for  20 
years,  at  4  per  cent  interest. 

By  (1)  of  Art.  217,  P  =  O-04)20-1  X      fi0°  7n  =  $6787. 
J  v   '  .04  (1.04)'20 

3.  Find  the  present  value  of  an  annuity  of  $1000  to  com- 
mence at  the  end  of  7  years  and  to  continue  14  years,  at  5 
per  cent  interest. 

By  (1)  of  Art.  218,  P  =       A     lR"  ~  ^ 
J  K  J  B  -  1\  Rp  +  n  J 

=  1000        (1.05)"  -  1  =  S7030> 
.05  (1.05)21 

4.  The  reversion  after  6  years  of  a  freehold  estate  is 
bought  for  $20000  ;  what  rent  ought  the  purchaser  to  receive, 
at  5  per  cent  compound  imterest?  Given  log  1.05  =  . 0211893, 
log  1.340096  =  .1271358. 

The  rent  is  equal  to  the  annual  value  of  the  perpetuity, 
deferred  for  6  years,  which  may  be  purchased  for  $20000. 

Let  A  be  the  value  of  the  annuity  in  dollars.  Then  by 
(2)  of  Art.  218, 

20000  =  ^M)-\ 
.05 

.-.     .4(1.05)-°  =  1000;  log.1  -  6  log  1.05  =  3. 
log^l  =  3.1271358  =  log  1340.096. 
.-.     A  =  $1340.090,  which  is  the  rent. 


456  EXAMPLES. 


EXAMPLES. 


1.    Express  log  y/a-4^3  in  terms  of  log  a  and  log&. 

Ans.  —  |  log  a  -f-  -jlogfr. 

2-   Simplify  log  [(£!)-  +  (^3)S].  -5  logo. 

3.  Simplify  log  V  729  V9-1  x  27~i  log  3. 

4.  Simplify  logff  -  21ogf  +  log^.  log  2. 

5.  Show  that  log  .V5,'  V2_  =  1  W  5  -  —  log  2  -  2  log  3. 

V18.V2      4  15  3 

Given  log 2  =. 3010300, log3=. 4771213, log 7  =  .8450980: 
find  the  value  of 

6.  log  84.  Ans.  1.9242793. 

7.  logy/ff.  .0563520. 

8.  log  14.4.  1.1583626. 

9.  log  [V  48  (108)*  -j-  ^6~].  1.0039238. 

10.  Given  log  2,  log  3,  log  7;  also  log  44092388 =7.6  143636, 

and   log  194.8445  =  2.2896883:    find  (1)   the   7th  root  of 
.00324;  and  (2)  the  11th  root  of  (39.2)2. 

Ans.  (1)  .44092388,  (2)  1.948445. 

11.  Given  log 2,  log3,  log 7;  alsolog9076. 226  =  3.9579053  : 
find  to  6  decimal  places  the  value  of  V[(294  x  'l25)-s-(42  x  32)]a. 

-      Ans.  9.076226. 

12.  Find  the  product  of  37.20:;,  3.7203,  .0037203,  37203, 
having  givdh  log  37.203  =  1.5705780,  and  log  1915631 
=  6.2823120.  Ans.  19156.31. 

13.  Find  the  number  of  digits  in  3U  x  28.  9. 

14.  Determine  the  number  of  ciphers  there  arc  between 
the  decimal  point  and  the  first  significant  digit  in  (.\)1000. 

Ans.  301, 

15.  Show  that  (21  ~  20)100>  100. 


EXAMPLES.  457 

Solve  the  following  equations  : 

a  log  a  -f-  log  6 

16.    ax  +  1  -=-  bx~x  =  c2*.         Ans.  — s__T — &_ — 

21ogc  —  log  a  +  log?; 

log  (^2  +  1) 


17.  2ato  +  ^  =  a-  21oga 

18.  a?  =  y* ,  a;3  =  2/2.  K  =  2i,  y  =  3| 

19.  a8*- ft89  =  m5,  x  _  4  log m       __      log m 
a3*  ■  ft2"  =  m10.  log  a  '  '  log  b 

».«•-•.*■-«■+■.*■.  iog610-"0g« 

■21.   <«e-W+i«)~»-(.-»)-(«  +  »)-.     ||g=g. 

Find  the  logarithms  of 

22.  .0625  to  base  2,  and  1000  to  base  .01.    ^4ws.  -4,  -f. 

23.  .0001  to  base  .001,  and  .1  to  base  90$.  |,  — |. 

24.  Va1',  1,  Vtf-V  to  base  a.  |,  -|,  -f. 

a* 

25.  Find  the  value  of  log^^  and  log34349.  — -|,  f. 

Solve  the  following  equations,  having  given  log  2,  log  3, 
and  log  7  : 

26.  2X-  6*-2  =  521  •  V~x.  Ans.  4.562. 

27.  2*+»  =6%        ^  =  log  3         .      =         log  2 

3*  =  3-2y  +  1.  log  3  —  log  2'  "        log  3  —  log  2' 

28.  31-T-"=4-y, 

3  log  3  -  2  log  2  log  3 

4(log3  -  log  2)'  J       4(log3  -  log  2) 

29.  Given  log10  2  =  .30103;  find  log26  200.  1.6465. 

30.  Given  log10  2  =  .30103,  log107  =  .84509  ;  find  log,^ 
and  logV27.  Ans   _log2.  =    1?81     2]og_7  =  ^6U, 

2  log  7  log  2 


458  EXAMPLES. 

81.  Given  M  =  .43429448,  and  log  2  =  .3010800;  find 
the  common  logarithms  of  7,  11,  17,  and  31  by  (1)  of 
Art.  210. 

Arts.  .8450980;  1.0413927;  1.2304489;  1.4913017. 

32.  Given  M and  log102  :  find  log101025.     Ans.  3.0107239. 
When  required,  the  following  logarithms   may  be  used  : 

log  2,  log  3,  log  7  (See  Ex.  6)  ;  also  log  11  =  1.0113927. 

33.  At  simple  interest,  the  interest  on  a  certain  sum  is 
$90,  and  the  discount  on  the  same  sum  for  the  same  time 
and  at  the  same  rate  is  $80  :  find  the  sum.  Ans.  720. 

34.  A  tradesman  marks  his  goods  with  two  prices,  one  for 
ready  money,  and  the  other  for  a  credit  of  G  months  :  find 
what  ratio  the  two  prices  ought  to  bear  to  each  other,  at  5 
per  cent  simple  interest.  Ans.  40  :  41. 

35.  Find  the  amount  of  $100  in  50  years  at  5  per  cent 
compound  interest ;  given  log  114.074  =  2.0594G50. 

Ans.  $1140.74. 

36.  In  how  many  years  will  a  sum  of  money  double  itself 
at  5  per  cent  compound  interest?  Ans.  14.2. 

37.  Find  the  present  value  of  $10000  due  8  years  heuce  at 
5  percent  compound  interest ;  given  log67683.94  =  4.83048,")6. 

Ans.  $6768. 39|. 

38.  In  how  many  years  will  $1000  become  $2500  at  10 
per  cent  compound  interest?  Ans.  9.6  years. 

39.  Show  that  money  will  increase  more  than  a  hundred- 
fold in  a  century,  at  5  per  cent  compound  interest. 

40.  What  sum  of  money  at  <">  per  cent  compound  interest 
will  amount  to  $1000  in  12  years?  Given  log  106=2.0253059, 
log  49697  =  4.6963292.    "  Ans.  $496.97. 

41.  In  how  many  years  will  a  sum  of  mone}'  treble  itself 
at  3£  per  cent  compound  interest?  Given  log  10850  = 
4.01494.  Ana.  Nearly  82. 

42.  A  person  borrows  $672  to  be  repaid  in  5  years  by 
annual  instalments  of  $120:  find  the  rate  of  interest  at  simple 

interest.  ^ins.  6  per  cent. 


EXAMPLES.  459 

43.  A  person  borrows  $600.25  :  find  how  much  he  must 
pay  annually  that  the  whole  debt  may  be  discharged  in  35 
years,  allowing  simple  interest  at  4  per  cent.       Arts.  $24.50. 

44.  Find  the  amount  of  an  annuity  of  $100  in  20  years, 
at  4|  per  cent  compound  interest.  Ans.  $3137.11. 

45.  A  freehold  estate  worth  $100  a  year  is  sold  for  $2500  : 
find  the  rate  of  interest.*  Ans.  4  per  cent. 

46.  The  reversion,  after  2  years,  of  a  freehold  worth 
$168.10  a  year  is  to  be  sold :  find  its  present  value  at  1\  per 
cent.  Ans.  $6400. 

47.  If  20  years'  purchase  must  be  paid  for  an  annuity  to 
continue  a  certain  number  of  years,  and  26  years'  purchase 
for  an  annuity  to  continue  twice  as  long ;  find  the  rate  per 
cent.  Ans.  3^. 

48.  When  4  per  cent  is  the  rate  of  interest,  find  what  sum 
must  be  paid  now  to  receive  a  freehold  estate  of  $400  a 
year  10  years  hence;  having  given  log  104  =  2.0170333, 
log  6.75565  =  .8296670.  Ans.  $6755.65. 

49.  If  a  perpetual  annuity  is  worth  20  years'  purchase, 
find  the  annuity  to  continue  for  3  years  which  can  be  pur- 
chased for  $2522.  Ans.  $800. 

50.  If  a  perpetual  annuity  is  worth  25  years'  purchase, 
find  the  amount  of  an  annuity  of  $625  to  continue  for  2 
years.  Ans.  $1275. 

*  In  these  examples  the  interest  is  compound  unless  otherwise  stated. 


460  IMAGINARY   QUANTITIES  —DEFINITIONS. 


CHAPTER    XXII. 

IMAGINARY    QUANTITIES.  — INDETERMINATE 
EQUATIONS.— THEORY     OF     NUMBERS. 

IMAGINARY    QUANTITIES. 

219.  Definitions.  —  An  Imaginary  Quantity  is  an  indi- 
cated even  root  of  a  negative  quantity.*     Thus, 

yda,    V-«*>     2  +  V^"^     3  ~  V"^     etc- 
are   imaginary  quantities.     They  are  also  called  impossible 
or  unreal  quantities. 

Since  no  number  whatever  can  have  a  negative  square,  it 
follows  that  if  we  have  to  extract  the  square  root  of  a  nega- 
tive quantity,  no  answer  is  possible,  in  ordinary  positive  or 
negative  numbers  (Art.  107).  But  although  the  square  root 
of  a  negative  quantity  is  thus  the  symbol  of  an  impossible 
operation,  yet  these  square  roots  frequently  occur  in  mathe- 
matical investigations,  and  it  becomes  important  to  explain 
in  what  sense  such  roots  are  to  be  regarded,  and  how  to  deal 
with  them. 

Every  imaginary  quantity  may  be  reduced  to  one  common 
form,  which  is  a  real  quantity  multiplied  by  the  imaginary 
quantity  V^— T-     Thus, 

Imaginary  quantities  may  be  added,  subtracted,  and  di- 
vided, the  same  as  other  surds;  but  when  we  attempt  to 
multiply  them  together  there  occurs  one  point  of  importance 
which  deserves  notice.  Thus,  by  the  rale  for  multiplying 
surds  together  (Art.  128),  we  have 

s/~i  x  v^  =  n/( -«■)(-«)  =  tf*5  =  ±«- 


All  other  quantities,  whether  rational  or  irrational,  are  ealled  real  quantities. 


FUNDAMENTAL   PRINCIPLES.  461 

But  since  the  square  root  of  any  number  multiplied  by  itself 
must,  by  the  definition,  give  the  original  number;  therefore 
y/— a  x  V— a  =  —a. 

.Similarly, 
y/-a  •  >f-b  =  \/a  •  \^-l  ■  \Jb  .  V^l  =  ^d)(sj~iy  =  -\fab~. 

The  surd  V  — 1  is  called  the  imaginary  unit.  The  imaginary 
unit  is  often  expressed  by  the  letter  i ;  but  until  the  student 
has  had  a  little  practice  in  the  use  of  imaginary  quantities, 
he  will  find  it  easier  to  retain  the  symbol  V—  1.  It  is  useful 
to  notice  the  successive  powers  of  V'— 1  or  i.  Thus,  since 
the  imaginary  unit  is  such  a  quantity  as  when  squared,  the 
result  is  —1,  therefore 

i   =    V^l, 

i»=  (V^l)2  =  -1, 

r»  =  (V/-1)3  =  (\^l)2.\d   =  -\^i, 

?4  =  (v^)4  =  (v7-!)2-  (v^T)2  =  (-i)(-i)  =  i, 

and  so  on,  repeating  the  results,  \f—  1,  —  1,  —  \] —  1,  1. 

The  sum  of  a  real  and  an  imaginary  quantity  is  called  a 
Complex  Quantity. 

It  is  usual  to  apply  the  term  imaginary  to  all  expressions 
which  arc  not  wholly  real.  Thus,  a  +  b\J  —\  may  be  taken 
as  the  general  type  of  all  imaginary  expressions.  Here  a 
and  b  are  real  quantities,  but  not  necessarily  rational. 

220.  Fundamental  Principles.  —  ( 1 )  If  a+b\J~\  =  0, 
then  a  =  0,  and  b  =  0. 

For,  if  a  +  ?V— 1  =  0, 

then     ZV— 1  =  -a;         .-.     -b2  =  a2;         .-.    a2  +  b2  =  0. 

Now  a2  and  b2  are  both  positive  ;  therefore  their  sum 
cannot  be  zero  unless  each  of  them  is  separately  zero  ;  i.e., 
a  =  05  and  b  =  0. 


462  FUNDAMENTAL   PRINCIPLES. 

(2)  If  a  +  b^l  =  c  +  dtf—L,  then  a  =  c,  and  b  =  d. 

For,  by  transposition,  a  —  c  +  (6  —  d)\J  —  1  =  0  ; 
therefore  by  (1),  a  —  c  =  0,  and  b  —  d  =  0;   i.e.,  a  =  c, 
and  &  =  d. 

Thus,  in  order  that  tivo  imaginary  expressions  may  be  equal, 
it  is  necessary  and  sufficient  that  the  real  parts  should  be 
equal,  and  the  imaginary  parts  should  be  equal. 

When  two  imaginary  expressions  differ  only  in  the  sign  of 
the  imaginary  part,  they  are  said  to  be  conjugate. 

Thus,  a  —  &V— 1  is  conjugate  to  a  +  b\l  —  1. 

The  sum  and  the  product  of  tivo  conjugate  imaginary 
expressions  are  both  reed. 

For,  a  +  b\l~\  +  a  -  b>T^i  =  2a    .     .     .     .   (1) 

Also,       (a  +  b)J~l)(a  -  Vf^X)  =  a"  +  b2      .     .    (2) 

The  use  of  imaginary  units  enables  us  to  factor  expressions 
which  are  prime  when  only  real  factors  are  admitted.  Thus, 
(2)  shows  that  the  sum  of  two  squares  can  always  be 
expressed  as  the  product  of  two  complex  factors. 

The  positive  value  of  the  square  root  of  a-  +  &2  is  called 
the  modulus  of  each  of  the  conjugate  expressions, 

a  +  6^1     and    a  -  &V^— 1. 

Divide  v'^a2  by  V^^. 

TT       ./ :.       j — To       \l— a2       rV-1       a 

Here  V— «2  -*■  V  —  b-  —    ,          =      , =s  -. 

\J-b2      by/ -i      h 

Hence  imaginary  expressions  are  divided  the  same  as  other 
surds. 

221.  Geometric  Meaning  of  the  Imaginary  Unit.  —  It  is 
clear  that  V— 1>  or  i,  cannot  represent  a  number,  or  any  real  quantity 
such  as  the  symbols  a,  b,  c,  etc.  represent;  it  must  signify  something 
which  a  real  quantity  does  not  take  account  of.  As  it  occurs  only 
as  a  multiplier  of  one  of  these  other  symbols,  it  must  denote  sonic 
operation  performed  on  the  real  quantity  denoted  by  these  symbols; 
and  the  result  of  operating  with  it  must  be  entirely  different  from 
that  of  operating  with  a  real  quantity. 


EXAMPLES.  463 

All  real  numbers  may  be  represented  in  magnitude  and  direction  by 
a  horizontal  line,  the  positive  numbers  increasing  in  one  direction 
from  zero,  and  the  negative  ones  in  the  opposite  direction  from  the 
same  point  (Art.  20). 

Now  ia  must  denote  the  result  of  some  operation  performed  on  a, 
and  i2a  must  similarly  denote  the  result  of  performing  this  operation 
twice. 

But  i2a  =  —a  (Art.  219).  Hence,  to  multiply  the  line  +a  twice  by 
the  imaginary  unit  i  reverses  the  sign  of  a.  We  have  seen  that  if  a 
denote  a  line  measured  in  one  direction,  — o  denotes  a  line  of  equal 
length  measured  in  the  opposite  direction,  a  line  which  could  be 
obtained  by  rotating  the  first  line  in  any  plane  through  two  right 
angles.     Hence, 

Multiplying  by  the  imaginary  unit  i  may  be  taken  to  denote  the 
operation  of  turning  the  line  a  through  one  right  angle,  in  any  plane 
which  contains  Us  original  iwsition. 

Thus,  let  BB'  be  drawn  perpendicular  to  AA' 
at  its  middle  point  0.  Denote  OA  by  the  symbol 
a  ;  then  multiplying  a  by  the  imaginary  unit  i 
turns  the  line  OA  through  the  right  angle  into 
the  position  OB.  Multiplying  again  by  i  turns 
the  line  through  the  second  right  angle  into  the 
position  OA',  opposite  OA.  Multiplying  a  third 
time  by  i  brings  the  line  into  the  position  OB1, 
opposite  OB.  Multiplying  by  i  a  fourth  time  restores  the  line  to  its 
original  position  OA.    That  is, 

OA  =  a;    OB  =  ia;     OA'  =  i2a  =  -a; 
OB'  =  i3a  =  —ia;    OA  =  i4a  =  a; 
which  corresponds  to  the  law  governing  the  powers  of  i  (Art.  219). 

Note.  —  This  is  but  a  brief  sketch  of  Imaginary  Quantities.  The 
subject  might  be  extended  to  great  length,  but  this  would  clearly 
be  beyond  the  limits  of  this  work.  The  student  who  wishes  to  pur- 
sue the  subject  further  is  referred  to  Peacock's  Algebra,  Warren's 
Treatise  on  the  square  root  of  negative  quantities,  Price's  Calculus, 
Vol.  I.  etc. 

EXAMPLES. 

Multiply  the  following : 

1.  2V^9  and  S^l. 

2.  24  +  V-49  and  24  -  ^-49. 

3.  2V^  -  6V^5  and  4\/3  -  ^/^~5. 

4.  Divide  6V^3  by  2V/^4. 


40  1  INDETERMINATE   /CQUATJONS. 


Express  with  rational  denominator: 

1 

11 

3  -  Sl-'l 

3V/-2  -  >l\P-b 

-19  -  6^10. 

7     3  +  5i 

-9  +  19* 

13 


INDETERMINATE    EQUATIONS. 


222.  Indeterminate  Equations.  —  We  have  seen  (Art. 
93),  that  the  definite  determination  of  the  values  of  two 
unknown  quantities  requires  two  independent  equations,  and 
that  when  only  one  equation  is  given  involving  two  unknown 
quantities,  there  will  generally  be  an  infinite  number  of  solu- 
tions ;  also,  if  there  be  any  number  of  independent  equa- 
tions involving  more  than  the  same  number  of  unknown 
quantities,  there  will  in  general  be  an  unlimited  number  of 
solutions.  Such  equations  are  called  indeterminate  equations. 
If  however  the  conditions  of  the  problem  be  such  as  to 
exclude  all  solutions  except  positive  integers,  the  number  of 
solutions  may  be  limited.  We  shall  discuss  only  the  simplest 
kinds  of  indeterminate  equations,  confining  our  attention  to 
'positive  integral  values  of  the  unknown  quantities  ;  the  others 
being  more  tedious  than  elegant,  and  of  not  much  practical 
use. 

Any  equation  of  the  first  degree  involving  two  unknowns, 
x  and  y,  can  be  reduced  to  the  form  ax  ±  by  =  ±c,  where 
a,  b,  c  are  positive  integers.  It  is  clear  that  the  equation 
ax  +  by  =  —  c  has  no  positive  integral  solution  ;  and  that 
the  equation  ax  —  by  =  — c  is  equivalent  to  by  —  ax  =  c ; 
hence  it  will  be  sufficient  to  consider  the  equations  ax±by=c. 

Neither  of  the  equations  ax  ±  by  =  c  can  be  satisfied  by 
integral  values  of  x  and  y  if  a  and  b  have  a  factor  which 
i Iocs  not  divide  c. 


INDETERMINATE   EQUATIONS.  465 

For,  if  either  equation  has  such  a  solution,  then,  dividing 
both  members  by  the  common  factor,  the  first  member  is 
integral  and  the  second  fractional,  which  is  impossible. 

Thus,  neither  of  the  equations  6a;  ±  9y  =  11  can  be 
solved  in  integers,  because  the  sum  or  difference  of  6a;  and 
9y  must  be  divisible  by  3  whatever  values  x  and  y  have. 

If  a,  6,  c  have  a  common  factor,  it  can  be  removed  by 
division,  so  that  in  the  examples  of  this  Article  we  shall 
suppose  that  a  and  b  are  prime  to  each  other. 

1.    Solve  7a;  +  12?/  =  220  in  positive  integers.      .     .   (1) 

Divide  by  7,  the  smaller  coefficient ;  thus, 

x  +  v  +  |'  =  31  +  f ; 

•  ••     x  +  y  +    y  =  PI. 

Since  x  and  y  are  to  be  integers, 

^zl  =  integer (2) 

7 

Multiplying  (2)  by  3  in  order  to  make  the  coefficient  of 
y  differ  by  unity  from  a  multiple  of  7,  we  have 


15y  -  9  ^  2y       1  +  V  -  2  -  integer. 

7                J                     7 

.*.     ^  —      =  integer  —  p,  suppose. 

.-.     y  —  2  =  7p;     or    y  =  7j>  +  2.     . 

.    (3) 

Substituting  in  (1),                           a;  =  28  -  12^.   ' 

•   (4) 

This  is  called  the  general  solution  of  the  equation  ;  and  by 
giving  to  p  any  integral  value,  we  obtain  from  (3)  and  (4) 
corresponding  integral  values  of  x  and  y  ;  but  if  p  >  2,  we 
see  from  (4)  that  x  is  negative  ;  and  if  p  is  a  negative  inte- 
ger, we  see  from  (3)  that  y  is  negative.     Thus,  the  only 


466  INDETERMINATE   EQUATIONS. 

positive  integral  values  of  x  and  y  are  obtained  by  putting 
p  =  0,  1,  2.     Thus  we  have 

p  =  0,  1,  2 
a;  =  28,  16,  4 
y  =    2,    1),  16 

Note.  —  An  artifice  similar  to  the  one  used  with  (2)  should  always 
be  employed  hefore  introducing  a  symbol  to  denote  the  integer. 

2.    Solve  5x  —  ly  =  29  in  positive  integers.     .     .     .   (1) 
Divide  by  5,  the  smaller  coefficient ;  thus, 
z  -  2/  -  f.'/  =  5  +  £ 


k        2y  +  4        .  , 
x  —  y  —  5  =  -^ — ! —  =  integer. 
5 


Multiplying  by  3, 


6v  +  12  .    „   .   y  +  2 

-       =  '/  +  2  +  •'  ^      =  integer. 

5  5 

V  +  2        .  4 
•  \     « — —  =  integer  =  p,  suppose, 
5 

y/  =  <sp  —  2; 

and  from  (1),  x  =  7p  +  3  | 

By  giving  to  p  any  positive  integral  value  we  obtain  posi- 
tive integral  values  of  x  and  y  ;  thus  we  have 

P  =     1,    2,    3,    4, 1 

x  =  10,  17,  24,  31, 

y  =     3,     8,  13,  18, J 

the  number  of  solutions  being  infinite. 

3.    In  how  many  ways  can  £500  be  paid  in  guineas  and 
live-pound  notes? 

Let  x  =  the  number  of  guineas, 
and     y  =  the  number  of  live-pound  notes. 

Then,  expressing  all  in  shillings,  we  have 

21»  +  1(%  =  10000. 


INDETERMINATE   EQUATIONS.  467 

Dividing  by  21,  x  +  Ay  +  %%y  =  476  +  /r. 

.-.     x  +  4?/  —  476  =  — — =  integer. 

J  21  ° 

16  -  G4g  =  _„  16^-j,  =  ^ 

21  J  21 

.•.     ^-^  =  integer  =  jp,  suppose. 

.-.  y  a  16  -  21p  ;  and  x  =  400  -f  lOOp. 
We  thus  see  that  y  is  negative  for  all  positive  values  of  p, 
and  that  if  p  is  negative  and  >  4,  x  is  negative.  Thus  the 
only  positive  integral  values  of  x  and  y  are  obtained  by 
putting p  =  0,  —1,  —2,  —3  ;  therefore  the  number  of  ways 
is  4.  If,  however,  the  sum  may  be  paid  either  in  guineas  or 
five-pound  notes,  p  may  also  have  the  value  —4.  If  p  =  —  4. 
then  x  =  0  and  y  =  100,  aud  the  sum  is  paid  entirely  in  five- 
pound  notes.  Thus,  if  the  zero  value  of  x  is  admissible, 
the  number  of  ways  is  5. 

4.  The  expenses  of  a  party  numbering  43  were  $114.50  ; 
if  each  man  paid  $5,  each  woman  $2.50,  and  each  child  $1, 
how  many  were  there  of  each? 

Let  »,  y,  z  denote  the  number  of  men,  women,  and 
children,  respectively  ;  then  we  have 

x  +  y  +  2  =  43 (1) 

5x  +  2.5//  +  z  =  114.5. 
Eliminating  z,  we  obtain  8x  -\-  3y  =  143. 
The  general  solution  of  this  equation  is 

x  =  Sp  +  1 ;     y  =  45  -  8p. 
Substituting  in  (1),  we  obtain  z  =  5p  —  3. 
Hence  p  cannot  be  negative  or  zero,  but  may  have  positive 
integral  values  from  1  to  5.     Thus  we  have 
p=     1,    2,    3,    4,    5; 
x  =     4,     7,  10,  13,  16; 
y  =  37,  29,  21,  13,    5; 
z  =     2,     7,  12,  1%  22. 


468  TI1E0RY  OF  NUMBERS. 

Solve  in  positive  integers : 

5.  3s+8#=103.     Ans.  3=29,  21, 13,  5;  y=2,  5,  8, 11. 

6.  5a; +2?/=  53.     sal,  3,  5,  7,  9  ;  #  =  24,  19,  14,  9,  4. 

7.  8a?  +  65y  =  81.  #        x  =  2;  y  =  1. 

8.  A  farmer  spends  $3760  in  buying  horses  and  cows ;  if 
each  horse  costs  $185  and  each  cow  $115,  how  many  of  each 
does  he  buy?  Ans.  11  horses,  15  cows. 

9.  In  how  many  ways  can  $100  be  paid  in  $1  pieces  and 
half-dollar  pieces,  including  zero  solutions?  Ans.  101. 

THEORY    OF    NUMBERS. 

223.  Scales  of  Notation.  —  In  the  ordinary  method  of 
expressing  numbers  *  1)3'  figures,  the  number  represented  by 
each  figure  is  always  a  multiple  of  some  power  of  ten.  For 
instance,  in  256  the  2  represents  2  hundreds,  i.e.,  2  times 
10'-;  the  5  represents  5  tens,  i.e.,  5  times  101 ;  and  the  6 
represents  6  units,  i.e.,  6  times  10°.  Thus,  4639  2«  an 
abbreviated  way  of  writing 

4  x  103  +  6  X  102  +  3  X  101  +  9. 

This  method  of  representing  numbers  is  called  the  common 
scale  of  notation,  and  10  is  said  to  be  the  base  or  the  radix 
of  the  scale.  The  figures  by  means  of  which  a  number  is 
expressed  are  called  digits.  The  symbols  employed  in  this 
system  of  notation  are  the  nine  digits  and  zero. 

In  like  manner  any  other  number  than  ten  may  be  taken 
as  the  radix  of  a  scale  of  notation  ;  thus  if  7  is  the  radix,  a 
number  expressed  by  3256  represents 

3x73  +  2x72  +  5x7  +  6; 

and  in  this  scale  no  digit  higher  than  6  can  occur. 

In  a  scale  whose  radix  is  denoted  by  r  the  above  number 
3256  represents  3r8  +  2r  +  5/-  +  6.  And  generally,  if  in 
a  scale  whose  radix  is  /•  the  digits,  beginning  with  that  in  the 

*  The  word  number  in  thto  subject  is  used  ;is  an  abbreviation  tor  positive  <"/<</"'• 


SCALES   OF  NOTATION.  46!) 

units'  place,  be  denoted  by  a0,  av  a2, a„,  then  the 

number  so  formed  will  be  represented  by 

anrn  +  an_!r"_1  -f-  n„_^'n_2  + +  <V'2  +  <V  +  ao- 

The  digits  an,  a„_l5 a0  are  integers,  each  is  less 

than  r,  and  any  one  or  more  of  them  after  the  first  may  be 
zero.  Hence  in  this  scale  the  number  of  digits,  including  0, 
is  equal  to  r,  and  their  values  range  from  0  to  r  —  1 . 

When  the  radix  is  2  the  scale  is  called  Binary;  when  3, 
Ternary;  when  10,  Denary,  or  Decimal;  when  12,  Duoden- 
ary or  Duodecimal;  etc.  The  most  useful  scale  after  the 
common  or  denary  is  the  duodenary.  Hence  it  is  necessary 
to  have  new  symbols  to  represent  the  digits  which  are  greater 
than  ten.  Usually  the  symbols,  t,  e,  T,  are  employed  as 
digits  to  denote  "ten,"  "eleven,"  and  "twelve."  Thus 
the  number 

6  t  7  e  4  in  the  scale  of  twelve  means 

6  x   124  +  10  x   123  +  7  x   12'2  +  11   x   12  +  4. 

It  is  not  usual  to  consider  any  scale  higher  than  the  duo- 
denary. 

Note.  —The  ordinary  operations  of  Arithmetic  may  be  performed 
in  any  other  scale  on  the  same  principle  as  in  the  common  scale.  In 
the  following  examples,  the  student  must  bear  in  mind  that  the  suc- 
cessive powers  of  the  radix  are  no  longer  powers  of  ten,  and  therefore, 
in  determining  the  carrying  figures,  he  must  not  divide  by  ten,  but 
by  the  radix  of  the  scale  in  use. 

1.  In  the  scale  of  eight  subtract  1732765  from  3673124, 
and  multiply  the  result  by  46. 


1740137 

3673124 

46 

1732765 

13501072 

1740137 

7600574 

111507032 

Explanation. —  Since  we  cannot  take  5  from  4  we  add  8  which 
makes  12;  then  5  from  12  leaves  7;  then  7  from  10,  which  leaves  3; 
then  8  from  9,  which  leaves  1;  and  so  on. 


470    TO  EXPRESS  A   NUMBER   IN  ANY  PROPOSED  SCALE. 

In  multiplying  by  6,  we  have 

6x7  =  forty-two  =  5  x  8  +  2; 

we  therefore  set  down  2  and  carry  5. 

Next  0x3  +  5  =  twenty-three  =  2  x  8  +  ?; 
set  down  7  and  carry  2;  and  so  on  till  the  multiplication  is  completed. 

In  the  addition,  4  -f-  7  =  eleven  =1  x  8  +  3;  we  therefore  set 
down  3  and  carry  1. 

Next  1  +  7  =  eight  =  1  x  8  +  0;  set  down  0  and  carry  1 ;  and  so 
on. 

2.  Divide  15  e  t  20  by  9  in  the  duodenary  scale. 

9)15  e  t  20 

1  e  e96 6 

Explanation.  — Since  15  =  1  x  T  -f  5  —  seventeen  =  1  x  9  +  8, 
we  set  down  1  and  carry  8. 

Next  8  x  T  +  e  =  one  hundred  and  seven  =  e  x  0  +  8;  we  there- 
fore set  down  e  and  carry  8;  and  so  on. 

3.  Find  the  square  root  of  300114  in  the  scale  of  5. 

300114(342 


114 


14 

1101 

1021 


1232 


3014 
3014 


4.  Add  23241,  4032,  300421  in  the  scale  of  five. 

Ans.  333244. 

5.  From  3  t  e  75G  take  2  e  46  t  2  in  the  duodenary  scale. 

Ans.  e  7074. 

6.  Multiply  6431  by  35  in  the  scale  of  seven.         334345. 

7.  Find  the  square  root  of  442641  in  the  scale  of  seven. 

Ans.  546. 

224.  To  Express  a  Whole  Number  in  Any  Pro- 
posed Scale. — 

Let  N  be  the  given  number,  r  the  radix  of  the  scale  in 
which  it  is  to  be  expressed,  and  a0,  av  <<.2, <<„  the 


EXAMPLES.  471 

required  digits  by  which  N  is  to  be  expressed  in  the   new 
scale,  beginning  with  that  in  the  units'  place  ;  then 

JSf  =  anrn  +  aH  _  ,  r"  ~ 1  + -f-  a./2  +  axr  +  a0 ; 

we  have  now  to  find  the  values  of  the  digits  a0,  av  a2,  .  .  .  an. 
Divide  N  by  r ;  then  the  remainder  is  a0,  and  the  quotient 
is 

anrn~l  +  a,^?-"-2  + +  a2r  +  av 

If  this  quotient  be  divided  by  r  the  remainder  is  av 

If  the  next  quotient  be  divided  by  r  the  remainder  is  a2 ; 
and  so  on,  until  there  is  no  further  quotient. 

Hence,  to  express  a  given  whole  number  in  any  proposed 
scale, 

Divide  the  number  by  the  radix,  then  the  quotient  by  the 
radix,  and  so  on  until  there  is  no  further  quotient ;  the  suc- 
cessive remainders  will  be  the  successive  digits  beginning  with 
the  one  in  the  units'  place. 

1.  Express  the  denary  number  5213  in  the  scale  of  seven, 
and  transform  21125  from  scale  seven  to  scale  eleven. 


(1)         7)5213 

7)744  ....  5 

7)106  ....  2 

7)15  ....  1 

2  ....  1 


(2)       e)21125 

e)1244  .  .  .  .  t 

e)61  .  .  .  .  o 

3  ....  t 


Explanation.  -  In  (1),  5213  =  2  x  74+  1  x  73+  1  x  7-+2  x  7  +  5; 
and  the  number  expressed  in  the  new  scale  is  21125. 

In  (2),  21  =  2  x  7  +  1  -  fifteen  =  1  X  c  +  4; 
therefore  on  dividing  by  e  we  set  down  1  and  carry  4. 

Next,  4x7  +  1=  twenty-nine  =  2  x  e  +  7 ; 
therefore  we  set  down  2  and  carry  7;  and  so  on. 

2.    Reduce  7215  from  scale  twelve  to  scale  ten  by  working 


472 


NUMBERS    IN    THE    COMMON   SYSTEM 


in  scale  ten,  and  tS47e  from  scale  twelve  to  scale  eleven  by 
working  in  scale  twelve. 


7215 

12 

e)£347e 

.  2 

e)e273  .  .  . 

(1) 

In  scale 
of  ten 

86 

12 

1033 

12 

e)102i  .  .  . 

eTlH  .  .  . 

e)12  .  .  . 

.  1 

2 

.  6 

.  3 

(2) 

In  scale 
of  twelve 

Result  is  12- 

12401 

01,  in  scale  of  ten. 

Result  is  136212 

in  scale  of  eleven. 

Explanation.  —  (1)  7215  in  the  scale  of  twelve  moans  7  x  123  +  2 
X  122  +  1  x  12  +  5  in  scale  ten.  The  calculation  is  most  easily  made 
by  writing  this  expression  in  the  form  j  (7  X  12  +  2)  x  12  + 1  \  x  12  +  5. 

(2)  Since  tS  —  t  x  T  +  S  =z  one  hundred  and  twenty-three  =  e 
X  e  -f  2,  we  set  down  e  and  carry  2. 

Also  2  x  T  +  4  =  twenty-eight  =  2  x  e  +  6; 

we  set  down  2  and  carry  6;  and  so  on. 

We  may  also  express  fractions  in  any  scale  of  notation;  thus, 


.356  in  scale  of  ten  denotes 
.356  in  scale  of  seven  denotes 
.356  in  scale  of  r  denotes 


i  +  A+A- 

10      102      103' 


1+ 


V 


3+ 


3.  Express  4954  in  the  scale  of  seven. 
Transform 

4.  212231  from  scale  four  to  scale  five. 

5.  23861  from  scale  nine  to  scale  eight. 

6.  333310  from  scale  ten  to  scale  eleven. 

7.  61520  from  scale  seven  to  scale  eleven. 


Ans.  20305. 

Am.  34402. 

37214. 

20846/. 

1105*. 


225.  Properties  of  Numbers  in  the  Common 
System.  —  A  number  which  has  no  factors  except  itself 
and  unity  is  called  a  prime  number,  or  a  prime;  a  number 
which  has  factors  besides  itself  and  unity  is  called  a 
composite  number;  thus.  .'11  is  a  prime  number,  and  L5  is 
a  composite  number.     Two  numbers  which  have  no  common 


NUMBERS   IN    THE   COMMON   SYSTEM.  4T6 

factor  except  unity  are  said  to  be  prime  to  each  other ;  thus, 
12  and  35  are  prime  to  each  other. 

(1)  If  a  number  p  divides  a  product  ab,  and  is  prime  to 
one  factor  a,  it  must  divide  the  other  factor  b. 

For,  since  p  divides  ab,  every  factor  of  p  is  found  in  ab ; 
but  since  p  is  prime  to  a,  no  factor  of  p  is  found  in  a ; 
therefore  all  the  factors  of  p  are  found  in  b ;  that  is,  p 
divides  b. 

(2)  If  a  prime  number  p  divides  a  product  abed , 

it  must  divide  one  of  the  factors  of  that  product. 

For  since  p  is  a  prime  number,  if  p  does  not  divide  a  it  is 
prime  to  a,  and  therefore  it  must  divide  bed  .  .  .  .  by  (1). 
Similarly,   if  p  does  not   divide  b   it   is    prime   to   b,   and 

therefore  it  must  divide  cd By  thus  proceeding  we 

shall  prove  that  p  must  divide  one  of  the  factors  of  the 
product. 

(3)  If  a  prime  number  p  divides  a",  where  n  is  any  positive 
integer,  it  must  divide  a. 

This  follows  directly  from  (2)  by  supposing  all  the  factors 
equal. 

(4)  If  p  is  prime  to  each  of  the  numbers  a  and  b,  it  is 
prime  to  their  product. 

For  since  p  is  prime  to  a,  and  b,  no  factor  of  p  can  divide 
a  and  b  ;  therefore  the  product  ab  is  not  divisible  by  any 
factor  of  p  ;  that  is,  p  is  prime  to  ab. 

Conversely,  if  p  is  prime  to  a&,  it  is  prime  to  each  of  the 
numbers  a  and  b. 

(5)  If  a  and.  b  are  prime  to  each  other,  am  and  b"  are  prime 
to  each  other ;  m  and  n  being  any  positive  integers. 

This  follows  at  once  from  (1);   hence  the  fractious  -  and 

—  are  in  their  lowest  terms. 
6" 

(6)  If  a  and  b  are  prime  to  each  other,  and  -  =  -,  then 

b       d 

c  and  d  must  be  equimultiples  of  a  and  b  respectively. 


474  THEOREMS   IN  RELATION   TO   NUMBERS. 

For,  if  -  =  -,  we  have  —  =  c ;  and  since  c  is  integral, 
—  is  integral ;    but  b  is  prime  to  a,  therefore  b  divides  d ; 

hence  d  =  nb,  where  n  is  some  integer ;  therefore  c  =  ?;a. 
226.    Theorems   in   Relation   to    Numbers.  —  The 

following  are  seven  theorems  of  some  importance  in  relation 
to  numbers : 

(1)  No  rational  Algebraic  formula  can  represent  prime 
numbers  only. 

For,  if  possible,  let  the  formula  a  +  bx  -+-  ex1  +  dx3 -j-  .  .  . 
represent  prime  numbers  only,  and  suppose  that  when  as  =  m 
the  value  of  the  expression  is  /»,  so  that 

p  =  a  -f  bm  +  cm2  +  dm3  -j-  .  .  . 

Now  let  x  =  m  +  np,  and  suppose  the  value  then  to  be  p'; 
that  is, 
p'  =  a  +  b(m  +  np)  +  c(m  +  np)2  -f-  d(m  +  vp)3  + 

=  a  +  bm  +  r?>r  +  dm3  + +  a  multiple  of  p 

=  p  -f  a  multiple  of  /> ; 
thus  the  expression  p'  is  divisible  by  p,  and  is  therefore  not 
a  prime  number. 

(2)  The  number  of  primes  is  infinite. 

For,  if  not,  let  p  be  the  greatest  prime  number ;  then  the 

product  2  •  3  •  5  •  7  •  11  • •  ]>  of  all  the  prime  numbers 

up  to  p,  is  divisible  by  each  of  these  prime  factors.  It 
follows  that  the  number  formed  by  adding  unity  to  this 
product  is  not  divisible  by  any  of  these  factors;  hence  this 
number  is  either  itself  a  prime  number,  or  it  is  divisible  by 
some  prime  number  greater  than  p  ;  thus  in  either  case  p  is 
not  the  greatest  prime  number.  Therefore  the  number  of 
primes  is  infinite. 

(3)  A  number  can  be  resolved  into  prime  factors  in  only 
one  icay. 

Let  N  denote  the  number;  suppose  X  =  abed  .  .  .  .  , 
where   a,   &,  c,  </, are  prime   numbers,   equal  or 


THEOREMS   IN  RELATION   TO   NUMBERS.  475 

unequal.   '  Suppose  also  that  N  =  a^c^ ,  where 

av   bv   cv   dv are  other  prime   numbers.      Then 

abed  .  .  .  =  ctj&jCjdj  .  .  .  ;  hence  at  must  divide  abed  .  .  . ; 
but  each  of  the  factors  of  this  product  is  a  prime  ;  hence  ax 
must  be  equal  to  one  of  them,  a  suppose.  Divide  by  ax  or  a  ; 
then  bed  .  .  .  .  =  bjCjC^  .  .  .  .  ;  as  before,  we  can  prove  that 

by  must  be  equal  to  one  of  the  factors  of  bed ;  and  so 

on.      Hence  the  factors  in  abed  ....  cannot  be  different 

from  those   in  a^b^d^ ,  and  therefore  N  can  be 

resolved  into  prime  factors  in  only  one  way. 

(4)    To  find  the  number  of  divisors  of  a  composite  number. 

Let  N  denote  the  number,  and  suppose  N  =  apbqcr  .  .  .  , 

where  a,  b,  c, are  different  prime  numbers,  and  p,  7, 

r,  ....  are  positive  integers.  Then  it  is  clear  that  the 
composite  number  is  divisible  by  each  of  the  numbers  1,  a, 

a2, ap,  1,  b,  b2, bq,  1,  c,  c2, cr, , 

and  also  by  the  product  of  any  two  or  more  of  them,  that  is, 
by  each  term  of  the  pi-oduct, 

(l+ft  +  ft2+    m    f   +fltP)(1  +  &+&2+   _   m    +6«)(1+C+C2+    _    m    +rr)    m    t  ? 

and  that  no  other  number  is  a  divisor  ;  hence  the  number  of 
divisors  is  the  number  of  terms  in  this  product,  that  is, 

(P+  !)(<?+  l)(r+  1) 


This  includes  among  the  divisors  both  unity  and  the  number 
itself. 

(5)  To  find  the  number  of  ways  in  which  a  composite  num- 
ber can  be  resolved  into  two  factors. 

Let  N  denote  the  number,  and  suppose  N  =  aPbqcr  .  .  .  .  , 

where  «,  6,  c, are  different  prime  numbers,  and  p,  7, 

r, are  positive  integers. 

First ;  suppose  N  not  a  perfect  square,  so  that  one  at  least 
of  the  exponents  p,  7,  r,  .  .  .  .  is  an  odd  number ;  then  the 
number  of  divisors,  by  (4),  is 

0>  +  1)(8  +  !*)('■+  1) 


476  THEOREMS   IN  RELATION   TO   NUMBERS. 

But  there  are  two  divisors  of  N  corresponding  to  each  way 
in  which  N  can  be  resolved  into  two  factors  ;  hence  the 
required  number  is 

Up  +  0(2  +  !)(*•+  i) 

Second  ;  suppose  N  a  perfect  square,  so  that  all  the  expo- 
nents p,  q,  r,  .  .  .  .  are  even.  In  this  case,  one  way  of 
resolution  into  factors  is  ViV  x  V7-^  and  to  this  way  there 
corresponds  only  one  divisor,  y/jSf.  If  we  exclude  this,  the 
number  of  ways  of  resolution  is 

*[(!»+  !)(?  +  !)(*■+  0 "l]i 

if  to  this  we  add  the  one  way  of  resolution,  ViV  X  v'iV,  we 
obtain  for  the  required  number, 

*[(P+  1)(9+  !)(*•+  1)  ....  +  !]• 

In  this  proposition  JV  X  1  is  counted  as  one  of  the  ways 
of  resolving  iVinto  two  factors. 

(G)  To  find  the  sum  of  the  divisors  of  a  number. 

Let  the  number  be   denoted   by  avbqcr as   before. 

Then  each  terra  of  the  product 

(l+a+cr+  .  .  +ap)(l+b+b*+  .  .  +6«)(l+c+c2+  .  .  +cr)  .  .  . 

is  a  divisor,  and  therefore  the  sum  of  the  divisors  is  equal  to 
this  product;  that  is  (Art.  163),  the  sum  required 

-  ((P  +  1  ~  l      °q+1  ~  l  .  r' H 1  -  1 
a  -  1  6-1  c  —  1 

(7)  T<>  find  the  number  of  ways  in  which  a  composite  num- 
ber can  be  resolved  into  two  factors  which  arc  prime  to  each 

other. 


THEOREMS   IN  RELATION   TO   NUMBERS.  All 

Let  the  number  N  =  apbncr as  before.     Since  the 

two  factors  are  to  be  prime  to  each  other,  we  cannot  have 
some  power  of  a  in  one  factor,  and  some  power  of  a  in  the 
other  factor,  but  ap  must  occur  in  one  of  the  factors.  Simi- 
larly bq  must  occur  in  one  of  the  factors  only ;  and  so  on. 
Hence  the  required  number  is  equal  to  the  number  of  ways 

in  which  the  product  abc can  be  resolved  into  two 

factors  ;  and  is  therefore,  by  (5), 

1(1  +  1)(1  +  1)0  +  1) ,or2-1, 

where  n  is  the  number  of  different  prime  factors  in  N. 

1.  Given  the  number  21  GOO,  find  (1)  the  number  of 
divisors,  (2)  the  sum  of  the  divisors,  and  (3)  the  number 
of  ways  in  which  it  can  be  resolved  into  two  factors  prime 
to  eacli  other. 

Smce  21600  =  63  -  10"2  =  25  •  33  •  5'2,  therefore 
by  (4) ,  the  number  of  divisors  =  (5  +  1 )  (3  +  1 )  (2  4- 1 )  =  72  ; 

26  _  l     o4  —  1     58  —  1 

6v  (6) ,  the  sum  of  the  divisors  = • • - 

JV/  2  —  13  —  lo  —  1 

=  63  x  40  x  31  =  78120; 

I  y  (7) ,  the  number  of  ways       =  23"1  =  4. 

2.  If  n  is  odd  show  that  n(n"  —  1)  is  divisible  by  24. 

We  have     w(n2  -  1)  =  n(n  -  l)(n  +  .1). 

Here  n  —  1,  n,  n  +  1  are  three  consecutive  numbers; 
hence  one  of  them  is  divisible  by  3  ;  and  since  n  is  odd, 
n  —  1  and  n  +  1  are  two  consecutive  even  numbers  ;  hence 
one  of  them  is  divisible  by  2  and  the  other  by  4.  Thus  the 
given  expression  is  divisible  by  the  product  of  2,  3,  and  4, 
i.e.,  by  24. 

3.  Find  the  least  multiplier  of  3675  which  will  make  the 
product  a  perfect  square.  Ans.  3. 

4.  Find  the  least  multiplier  of  1845  which  will  make  the 
product  a  perfect  cube.  Ans.  3  •  5a  •  41". 

5.  Find  the  number  of  divisors  of  140.  Ans.  12. 


478  EXAMPLES. 

EXAMPLES. 

Multiply 

!.    2\^3  +  3V^2  by  4\^3  -  5^2.     4ns.  6  -  2^0. 
2.    3y/~7  -  5\^2  by  2>\l~l  +  5^-1.  -13. 


3.    f~l  +  e"^1  by  e^"1 


+  V-3byx_L^B. 


2 

Express  with  rational  denominator 


5. 

3  +  2V^1    j    3  -  2^-1 
2  _  5^-1        2  +  5\/^T 

^             8 
vlns.  —  — . 
29 

6. 

X   -   V-l                 »   +   V-l 

2  (3a 

r  -  1)V^1 

x-  +  1 

(a  +  V^T)3  -  (a  -  V~T)3 

3«-  -  1. 

(a  +  S/-1)2-  (a-s/-l>" 

2a 

8. 

(i  +  o2 

3  -  i 

-1  -f  3t 
5 

9. 

\fe  -  /\/2 

8  -  ?V6 

2\ll  -  iV2~" 

14 

10. 

1  +  i 

i. 

1  -  j 

11.  Find  the  value  of  (—  V7  —  l)4'1"13,  when  ?i  is  a  positive 
integer.  -Ih.s.  V  —  1. 

12.  Find  the  square  of  ^9 +40^^1  +  ^9-40^^!.       100. 
Solve  in  positive  integers  : 

13.  7x  +  12y  =     152.  Ans.  x  =  20,  8  ;  y  =  1,  8. 

14.  13a;  +  11//  =  414.  x  =  9,  20,  31  ;  y  =  27,  14,  1. 

15.  23a  +  25y  =  915.  x  =  30,  5  ;  //  =  9,  32. 

16.  41a;  +  47//  =  2191.  a;  =  50,  3  ;  y  =  3,  1  I. 

17.  17a;  +  23?/  =  183.  x  =  4 ;  //  =  5. 

18.  19.r  +    5//  =  119.  a;  =  1,  6;  y  =  20,  1. 


EXAMPLES.  479 

19.  In  how  many  ways  can  $2000  be  paid  in  $21  bills  and 
$5  bills,  supposing  there  to  be  sueh  bills?        Ans.  19,  or  20. 

20.  In  how  many  ways  can  $4000  be  paid  in  $21  and  $40 
bills?  Ans.  4,  or  5. 

21.  In  how  many  ways  can  $39  be  paid  in  $4  and  $5  bills? 

Ans.  2. 

22.  Find  how  many  dollars  and  half-dollars,  whose  diam- 
eters are  respectively  .81  and  .666  of  an  inch,  may  be  placed 
in  a  row  together,  so  as  to  make  a  yard  in  length. 

Ans.  28,  20. 

23.  A  farmer  buys  oxen,  sheep,  and  ducks.  The  whole 
number  bought  is  100,  and  the  whole  sum  paid  is  £100. 
Supposing  the  oxen  to  cost  £5,  the  sheep  £1,  and  the  ducks 
Is.  per  head,  find  the  number  he  bought  of  each. 

Ans.  19  oxen,  1  sheep,  80  ducks. 

24.  I  buy  40  animals  consisting  of  sheep  at  $20,  pigs  at 
$10,  and  oxen  at  $85  ;  if  I  spend  $1505,  how  many  of  each 
do  I  buy? 

Ans.  28  sheep,  1  pig,  11  oxen  ;  or  13  sheep,  14  pigs,  13  oxen. 

25.  Add  303478,  150732,  264305  in  the  scale  of  nine. 

Ans.  728626. 

26.  Find  the  product  of  4685  and  3483  in  the  scale  of 
nine.  Ans.  17832126. 

27.  Divide  14332216  by  6541  in  the  scale  of  seven. 

Ans.  1456. 

28.  Divide  the  difference  between  121012  and  11022201 
by  1201  in  the  scale  of  three.  Ans.  2012. 

29.  Divide  102432  by  36  in  the  scale  of  seven.    Ans.  1625. 

30.  Divide  17832126  by  4685  in  the  scale  of  nine. 

Ans.  3483. 

31.  Extract  the  square  root  of  33224  in  the  scale  of  six. 

Ans.  152. 

32.  Extract  the  square  root  of  11000000100001  in  the 
scale  of  two.  Ans.  1101111. 

33.  Extract  the  square  root  of  67556^21  in  the  scale  of 
twelve.  Ans.  8te7. 


480  EXAMPLES. 

34.  Extract  the  square  root  of  e.etOOX  in  the  scale  of 
twelve.  Ans.  eee. 

35.  Express  624  in  the  scale  of  five.  114 1. 

36.  Express  1458  in  the  scale  of  three.  2000000. 
Transform 

37.  213014  from  scale  six  to  scale  nine.  Arts.  2G011. 

38.  400803  from  scale  nine  to  scale  five.  30034342. 

39.  123456  from  scale  ten  to  scale  seven.  1022634. 

40.  1357531  from  scale  ten  to  scale  live.  321420111. 

41.  15951  from  scale  eleven  to  scale  ten.  22441. 

42.  Find  the  least  multiplier  of  3234  which  will  make  the 
product  a  perfect  square.  Ans.  66. 

43.  Find  the  least  multiplier  of  6480  which  will  make  the 
product  a  perfect  cube.  Ans.  2'2  •  3'-  •  52. 

44.  Find  the  least  multiplier  of  13168  which  will  make  the 
product  a  perfect  cube.  2-  •  (.S2:3)J. 

45.  Find  the  least  multipliers  of  the  numbers  4374,  18375, 
74088,  respectively,  which  will  make  the  products  perfect 
squares.  Ans.  6,  15.  12. 

46.  Find  the  least  multipliers  of  the  numbers  7623, 
109350,  539539,  respectively,  which  will  make  the  products 
perfect  cubes.  Am.  1617,  180,  18.".:). 

47.  Find  the  number  of  divisors  of  8064.  Ans.  48. 

48.  In  how  many  ways  can  the  number  7056  be  resolved 
into  two  factors?  Ans.  22. 

49.  Find  the  number  of  divisor,s  of  1845";  and  the  num- 
ber of  ways  iu  which  it  can  be  resolved  into  two  factors. 

Am.  12;  6. 

50.  Find  the  number  of  divisors  there  are  in  |_9_,  and  the 
sum  of  these  divisors.  Ans.  160;  1181040. 

51.  Find  the  number  of  divisors  of  1000,  3600,  and 
14553.  Ans.  16,  45,  24. 

52.  In  how  many  ways  can  a  line  100R00  inches  long  lie 
divided  into  equal  parts,  each  some  multiple  of  an  inch? 

Ans.  126. 


PROBABILITY    {CHANCE)  —  DEFINITION.  481 


CHAPTER     XXIII. 

PROBABILITY     (CHANCE). 

227.  Definition.  —  If  an  event  can  happen  in  a  ways 
and  fail  in  b  ways,  and  if   each  of  these   ways  is  equally 

likely,  the  probability  or  chance  of  its  happening  is ; 

and  the  probability  or  chance  of  its  failing  is  -. 

a  -f  b 

Otherwise,  thus :  If  an  event  can  happen  in  a  ways  and 
fail  in  b  ways,  and  each  of  these  ways  is  equally  likely  to 
happen,  we  can  assert  that  the  chance  of  its  happening  is 
to  the  chance  of  its  failing  as  a  to  b  ;  therefore  the  chance 
of  its  happening  is  to  the  sum  of  the  chances  of  its  happen- 
ing and  failiug  as  a  to  a  +  b.  But  the  event  must  either 
happen  or  fail,  hence  the  sum  of  the  chances  of  its  happen- 
ing and  failing  must  represent  certainty.  Therefore  the 
chance  of  its  happening  is  to  certainty  as  a  to  a  +  b.  If 
then  we  represent  certainty  by  unity,  we  have 

the  chance  that  the  event  will  happen  is ; 

a  +  b 

and  the  chance  that  the  event  will  not  happen  is  -. 

a  +  b 

Hence  we  have  the  following  definition  : 

The  probability  of  an  event  happening  is  the  ratio  of  the 
number  of  favorable  cases  to  the  whole  number  of  cases  both 
favorable  and  unfavorable.  * 

Thus,  if  a  bag  contain  5  white  and  12  black  balls,  and  one 
of  them  be  drawn  at  random  from  the  bag,  the  chance  of 
this  being  a  white  ball  is  ^,  and  of  its  being  a  black  ball 

IS    TTT« 


482  PROBABILITY   (CHANCE)  —  DEFINITION. 

If  p  is  the  chance  of  the  happening  of  an  event,  the 
chance  of  its  not  happening  is  1  —  p. 

Note.  —  When  the  chance  of  the  happening  of  an  event  is  to  the 
chance  of  its  failing  as  a  to  b,  it  is  sometimes  expressed  in  popular 
language  thus:  the  odds  are  a  to  b  in  favor  of  the  event,  or  b  to  a 
against  the  event. 

If  the  chances  for  and  against  an  event  are  even,  the 
chance  of  the  event  is  %. 

1.  From  a  bag  containing  3  white  and  0  black  balls  a 
man  draws  one  at  random  :  what  is  the  chance  that  this  will 
be  a  white  ball? 

There  are  here  3  ways  of  drawing  a  white  ball  and  12 
ways  of  drawing  either  a  white  or  a  black  ball. 
Hence  the  required  chance  =  T%  =  J. 

2.  From  a  bag  containing  4  white  and  5  black  balls  a 
man  draws  3  at  random :  what  is  the  chance  of  these  being 
all  black? 

The  total  number  of  ways  in  which  3  balls  can  lie  drawn 
from  9  balls  is  9<78  (Art.  181),  and  the  number  of  ways  of 
drawing  3  black  balls  from  5  is  5C8 ;  therefore  the  chance 
of  drawing  3  black  balls 

=  &  =  t±il  1  A. 

9C3       9.8-7       42 

3.  Find  the  chance  of  throwing  at  least  one  ace  in  a  single 
throw  with  two  dice. 

As  a  common  die  has  6  symmetric  faces,  it  may  fall  in 
any  one  of  6  ways  ;  and  each  of  these  6  ways  may  unite 
with  each  of  the  six  ways  of  the  second  die,  giving  36 
different  cases  in  all.  An  ace  on  one  die  may  lie  associated 
with  any  of  the  G  numbers  on  the  other  die,  and  the 
remaining  5  numbers  on  the  first  die  may  each  be  associated 
with  the  ace  on  the  second  die  ;  thus  the  number  of  favorable 
eases  is  11. 

Hence  the  required  chance  =  ££. 


SIMPLE  EVENTS.  483 

Otherwise  thus  : 

There  are  5  ways  in  which  each  die  can  be  thrown  so  as 
not  to  give  an  ace  ;  hence  25  throws  of  the  two  dice  will 
exclude  aces,  so  that  the  chance  of  not  throwing  one  or  more 
aces  is  §§. 

Hence  the  chance  of  throwing  at  least  one  ace  =  1  —  |-| 

4.  Find  the  chance  of  throwing  more  than  15  in  one  throw 
with  3  dice. 

A  throw  amounting  to  18  must  be  made  up  of  Go,  6,  6,  and 
this  can  occur  in  1  way  ;  17  can  be  made  up  of  G,  G,  5, 
and  this  can  occur  in  3  ways  ;  16  can  be  made  up  of  G,  6, 
4,  and  G,  5,  5,  and  these  can  each  occur  in  3  waj's.  There- 
fore the  number  of  favorable  cases  is 

1  +  3  +  3  +  3,  or  10. 

And  the  whole  number  of  cases  is  63,  or  21G. 
Hence  the  required  chance  =  ^  =  T§¥. 

228.   Simple  Events.  —  Suppose  there  to  be  a  number 

of  events,  A,  B,  C, ,  of  which  one  must,  and  only 

one   can,   occur ;    and  suppose  a,   b,   c, to  be  the 

numbers  of  ways  respectively  in  which  these  events  can 
occur,  and  that  each  of  these  ways  is  equally  likely  to 
happen.  Then,  since  A  can  occur  in  a  ways,  out  of 
a  _|_  0  _|-  c  + ways,  therefore  by  Art.  227, 

the  chance  that  A  will  happen  is . 

a  +  b  +  c  .  .  . 

Similarly,  the  chance  that  B  will  happen  is — ■ ; 

J  a  +  6  +  c  .  .  . 

and  so  on. 

1.  From  a  bag  containing  3  white,  4  black,  and  5  red 
balls  a  man  draws  2  at  random  :  what  are  the  probabilities 
of  the  different  cases  ? 

The  number  of  pairs  that  can  be  formed  out  of  12  things 


484  SIMPLE   EVENTS. 

is  V2C2  (Art.  181),  or  66.     The  number  of  pairs  that  can  be 
formed  out  of  the  three  white  balls  is  8C2  or  3  ;  hence 

the  chance  of  drawing  two  white  balls  =  -fa. 
Similarly,  the  chance  of  drawing  two  black  balls  ==  &  ; 
and  the  chance  of  drawing  two  red  balls      =  £§. 

Also,  as  each  of  the  3  white  balls  might  be  associated  with 
each  of  the  4  black  balls,  the  number  of  pairs  consisting  of 
one  white  and  one  black  ball  is  3  X  4,  or  12  ;  hence 

the  chance  of  drawing  a  white  and  a  black  ball  =  £f . 
Similarly, 

the  chance  of  drawing  a  black  and  a  red  ball       =  §#; 

and  the  chance  of  drawing  a  red  and  a  white  ball      =  $f . 

The  sum  of  these  six  chances  is  unity,  as  it  should  be 
(Art.  227). 

2.  A  has  3  shares  in  a  lottery  in  which  there  are  3  prizes 
and  6  blanks  ;  B  has  1  share  in  a  lottery  in  which  there  are 
1  prize  and  2  blanks  :  show  that  A's  chance  of  success  is  to 
B's  as  16  to  7. 

A  may  draw  three  prizes  in  one  way  ;  he  may  draw  2 
prizes  and  1  blank  in  3C2  X  G,  or  18  ways  ;  he  may  draw 
1  prize  and  2  blanks  in  3  x  6C2,  or  45  ways  :  the  sum  of 
these  numbers  is  64,  which  is  the  number  of  ways  in  which 
A  can  draw  a  prize.  Also  he  can  draw  3  tickets  in  963,  or 
84  ways  ; 

.-.     A's  chance  of  drawing  a  prize  =  §£  =  -|f ; 

and  clearly  B's  chance  of  drawing  a  prize  =  £ ; 
therefore  A's  chance  :  B's  chance  : :  16  :  7. 

Or  ive  might  have  reasoned  thus :  A  may  draw  3  blanks  in 
6C8,  or  20  ways. 

.-.     A's  chance  of  drawing  a  blank  =  §£  =  ^  ; 
.-.     A's  chance  of  drawing  a  prize  =  1  —  7j5r  =  £-$-  (Art.  227) . 

3.  In  a  single  throw  with  two  dice  find  the  chance  (1)  of 
throwing  live,  and  (2)  of  throwing  six.     Ans.  (1)  J  ;  (2)  /'0. 


COMPOUND  EVENTS.  485 

4.  From  a  pack  of  52  cards  two  are  drawn  at  random  : 
find  the  chance  that  one  is  a  knave  and  the  other  a  queen. 

Ans.  ^. 

5.  Find  the  chance  of  drawing  two  black  halls  and  one 
red  hall  from  a  hag  containing  5  black,  3  red,  and  2  white 
halls.  Ans.  J. 

229.  Compound  Events.  —  Hitherto  we  have  consid- 
ered only  those  oecurrences  which  in  the  language  of  Proba- 
bility  are  called  Simple  events.  When  two  or  more  of  these 
simple  events  occur  in  connection  with  each  other,  the  joint 
occurrence  is  called  a  Compound  event.  Events  are  said  to 
be  dependent  or  independent  according  as  the  occurrence  of 
one  does  or  does  not  affect  the  occurrence  of  the  other. 
Dependent  events  are  sometimes  said  to  be  contingent. 

(1)  If  there  are  two  independent  events  of  which  the  re- 
spective probabilities  are  Jcnoivn,  to  find  the  probability  that 
both  will  happen. 

Suppose  that  the  first  event  may  happen  in  a  ways  and 
fail  in  b  ways,  all  these  ways  being  equally  likely  to  occur; 
and  suppose  that  the  second  event  may  happen  in  a'  ways 
and  fail  in  V  ways,  all  these  ways  being  equally  likely  to 
occur.  Each  of  the  a  +  b  cases  may  be  associated  with 
each  of  the  a'  +  b'  cases,  thus  forming  («  +  b)  (a'  +  b') 
compound  cases  all  equally  likely  to  occur. 

In  aa'  of  these  compound  cases  both  events  happen,  in  bb' 
of  them  both  events  fail,  in  ab'  of  them  the  first  event  hap- 
pens and  the  second  fails,  and  in  alb  of  them  the  first  event 
fails  and  the  second  happens.     Thus, 

. — is  the  chance  that  both  events  happen  ; 

(a  +  &)  («'  +  U) 

is  the  chance  that  both  events  fail ; 


(a  +  6)  (a'  +  i') 


is  th»  chance  that  the  first  happens  and 

(a  +  6)  (a'  +  &') 

the  second  fails. 


486  COMPOUND   EVENTS. 

—  is  the  chance  that  the  first  fails  and  the 

(a  +  b)  (a'  +  b') 

second  happens. 

Hence,  If  tioo  events  be  independent  of  each  other,  the 
probability  that  they  will  both  happen  is  equal  to  the  product 
of  their  separate  probabilities. 

Thus,  if  p  and  p'  are  the  respective  chances  of  two  in- 
dependent events,  pp'  is  the  chance  that  both  events  will 
happen.  Similar  reasoning  will  apply  in  the  case  of  any 
number  of  independent  events.  For  example,  suppose  there 
are  three  independent  events,  and  that  p,  p\  p"  are  the  re- 
spective chances  of  their  happening  separately.  From  the 
above  rule,  the  chance  that  the  first  two  will  happen  is  pp' ; 
then  in  the  same  way  the  chance  that  the  first  three  will 
happen  is  pp'  x  p",  or  pp'p"  ;  and  so  on  for  any  number. 
Similarly,  the  chance  that  all  the  events  will  fail  is 
(1  —  p)(l  —  j/)(l  —  p")-  The  chance  that  the  first  two 
events  will  happen  and  the  other  fail  is  pp'{l  —  p")  ;  and 
so  on. 

Hence,  If  there  be  any  number  of  independent  events,  the 
probability  that  they  will  all  happen  is  equal  to  the  product  of 
their  separate  probabilities. 

If  p  is  the  chance  that  an  event  will  happen  in  one  trial, 
the  chance  that  it  will  happen  in  any  assigned  succession  of 
r  trials  is  pr ;  this  follows  at  once  by  supposing  that  p,  p' , 
p" ', are  each  equal  top. 

1 .  Find  the  chance  of  throwing  an  ace  in  the  first  only  of 
two  successive  throws  with  a  single  die. 

Here  we  require  the  ace  to  appear  at  the  first  throw,  and 
at  the  second  throw  it  is  not  to  appear. 

The  chance  of  throwing  the  ace  at  the  first  trial  =  I  ;  and 
the  chance  of  not  throwing  the  ace  at  the  second  trial  =  |. 

.-.     the  chance  of  the  compound  event  =  $  K  f  =  -;[;;. 

2.  Suppose  5  white  and  8  black  balls  to  be  placed  in  a 
bag;   find  the  chance  that  in   two  successive  drawings  of  3 


COMPOUND   EVENTS.  487 

balls  each,  the  first  will  give  3  white  and  the  second  3  black 
balls,  the  balls  first  drawn,  being  replaced  before  the  second 
trial. 

The  number  of  ways  in  which  3  balls  may  be  drawn  is 

The  number  of  ways  in  which  3  white  may  be  drawn  is  6C8 ; 

The  number  of  ways  in  which  3  black  may  be  drawn  is  8C3. 

.-.     the  chance  of  3  white  at  the  first  trial  =  5C3-^13C3=T*£^  ; 

and  the  chance  of  3  black  at  the  second  trial =8C3-r-13C3=  ^fe. 

.-.     the  chance  of  the  compound  event  =  Tf  3  X  T2j3  =  ?hi%9- 

Note.  — To  find  the  chance  that  some  one  at  least  of  the  events 
will  happen  we  proceed  thus:  the  chance  that  all  the  events  fail  is 

(1  —p)(l  —  2/)(l  —  P") ,  and  except  in  this  case  some  one  of 

the  events  must  happen;  hence  the  required  chance  is 

1_(1  -  p)(l  -  p')U  -  P") 

3.  Suppose  that  it  is  9  to  7  against  a  person  A  who  is 
now  35  years  of  age  living  till  he  is  65,  and  3  to  2  against  a 
person  B  now  45  living  till  he  is  75  :  find  the  chance  that 
one  at  least  of  these  persons  will  be  alive  30  years  hence. 

The  chance  that  A  will  die  within  30  years  =  T% ; 
and  the  chance  that  B  will  die  within  30  years  =  f  ; 
. •.  the  chance  that  both  will  die  within  30  years  =  |  X  fo  =  §$■ 

Therefore  the  chance  that  both  will  not  be  dead,  that  is,  that 
one  at  least  will  be  alive,  is  1  —  f£  =  §§. 

(2)  The  probability  of  the  concurrence  of  two  dependent 
events  is  the  product  of  the  probability  of  the  first  into  the 
probability  that  when  that  has  happened  the  second  will 
follow. 

This  is  only  a  slight  modification  of  the  principle  estab- 
lished in  (1),  and  is  proved  in  the  same  way.  Thus,  suppose 
that  when  the  first  event  has  happened,  d  denotes  the  number 
of  ways  in  which  the  second  event  will  follow,  and  b'  the 
number  of  ways  in  which  it  will  not  follow  ;  then  the  number 


488  MUTUALLY   EXCLUSIVE   EVENTS. 

of  ways  in  which  the  two  events  can  happen  together  is  aa', 
and  the  probability  of  their  concurrence  is 


(a  +  b)  (a'  +  b') 
Thus,  if  p  is  the  probability  of  the  first  event,  and  p'  the 
contingent  probability  that  the  second  will  follow,  the  prob- 
ability of  the  concurrence  of  the  two  events  is  ppf. 

4.  In  a  hand  at  whist  find  the  chance  that  a  specified 
player  holds  both  the  king  and  queen  of  trumps. 

Denote  the  player  by  A  ;  then  the  chance  that  A  has  the 
king  is  clearly  13  -e-  52  ;  for  this  particular  card  can  be  dealt 
in  52  different  ways,  13  of  which  fall  to  A.  The  chance  that, 
when  A  has  the  king,  he  can  also  hold  the  queen  is  then 
12  -r-  51  ;  for  the  queen  can  be  dealt  in  51  ways,  12  of  which 
fall  to  A. 

Therefore  the  chance  required  =  if  X  Jf  =  TV 

5.  Suppose  5  white  and  8  black  balls  to  be  placed  in  a 
bag:  find  the  chance  that  in  two  successive  drawings  <>f  :) 
balls  each,  the  first  will  give  3  white  and  the  second  3  black 
balls,  the  balls  first  drawn  not  being  replaced  before  the  seconk 
trial. 

At  the  first  trial,  3  balls  may  be  drawn  in  13C3  ways  ;  and 
3  white  balls  may  be  drawn  in  5C3  ways. 

At  the  second  trial,  3  balls  may  be  drawn  in  l0C3  ways  ; 
and  3  black  balls  may  lie  drawn  in  8C3  ways. 
.-.     the  chance  of  3  white  at  first  trial       =  5C3  -=-  i3C3  =  T|^  ; 
and  the  chance  of  3  black  at  second  trial  =  8C3  -5-  13C3  =  ^-. 
the  chance  of  the  compound  event  =  Tiu  X  it  =  ila- 

The  student  should  compare  this  solution  with  that  in 
Ex.  2. 

230.  Mutually  Exclusive  Events.  —  If  an  event  can 
happen  in  two  or  more  different  ways  which  are  mutually 
exclusive,  tin'  chance  that  it  will  happen  is  the  sum  of  the 
chances  of  its  happening  in  these  different  ivays. 


MUTUALLY   EXCLUSIVE   EVENTS.  489 

If  the  different  ways  of  happening  are  all  equally  probable, 
this  proposition  is  merely  a  repetition  of  the  definition  of 
probability  (Art.  227)  ;  and  if  they  are  not  equally  probable, 
the  proposition  seems  to  follow  so  naturally  from  that  defi- 
nition that  it  is  often  assumed  as  self-evident.  It  may, 
however,  be  proved  as  follows  : 

Suppose  the  event  can  happen  in  two  ways  which  cannot 

concur;  and  let  -,    —  be  the  chances  of  the  happening  of 
b     b 

the  event  in  these  two  ways,  respectively.     Then  out  of  bb' 

cases  there  are  ab'  in  which  the  event  may  happen  in  the  first 

way,  and  a'b  in  which  the  event  may  happen  in  the  second 

way  ;  and  these  ways  cannot  concur.     Therefore,  out  of  the 

whole  bb'  cases  there  are  ab'  -f-  a'b  cases  favorable  to  the 

event ;  hence  the  chance  that  the  event  will  happen  in  one  or 

the  other  of  the  two  ways  is 

ab'  +  a'b  _  a   ,   a[ 

bb'        ~  b       b'' 

Similarly  for  any  number  of  exclusive  ways  in  which  the 
event  can  happen. 

Hence  if  an  event  can  happen  in  n  ways  which  are  mutu- 
ally exclusive,  and  if  pv  p2,  p8, pn  are  the  proba- 

bilities  that  the  event  will  happen  in  these  different  ways, 
respectively,  the  probability  that  it  will  happen  in  some  one 
of  these  ways  is 

Pi    +    Pi    +    P3    + +    P- 

1.  Suppose  two  urns,  A  and  B;  let  A  contain  2  white 
and  3  black  balls,  and  let  B  contain  3  white  and  4  black 
balls  ;  find  the  chance  of  obtaining  a  white  ball  by  a  single 
drawing  from  one  of  the  urns  taken  at  random. 

Since  each  urn  is  equally  likely  to  be  taken,  the  chance  of 
selecting  the  urn  A  is  ^  ;  and  the  chance  then  of  drawing 
a  white  ball  from  it  is  §  ;  hence  the  chance  of  obtaining  a 
white  ball  so  far  as  it  depends  on  A  is  \  X  f ,  or  \.     Simi- 


490  MUTUALLY   EXCLUSIVE  EVENTS. 

larly,   the   chance   of   obtaining  a  white  ball  so  far  as  it 
depends  on  B  is  |   x  -2,  or  T3T. 

.-.     the  required  chance  =  \  +  T\  =  f$. 

2.  Find  the  chance  of  throwing  9  at  least  in  a  single 
throw  with  two  dice. 

There  are  36  cases  in  all ;  9  can  be  made  up  in  4  ways,  10 
in  3  ways,  11  in  2  ways,  and  12  in  1  way. 

.-.     the  chance  of  throwing    9  =  4  -5-  36 ; 

the  chance  of  throwing  10  =  3  -f-  36  ; 

the  chance  of  throwing  11  =  2  -=-  36  ; 

and  the  chance  of  throwing  12  =  1  -f-  36. 

Now  the  chance  of  throwing  a  number  not  less  than  9  is 

the  sum  of  these  separate  chances. 

..  .     ,    ,  4  +  3  +  2  +  1         5 

.-.     the  required  chance  =  — ! ! — ■  =  — . 

36  18 

Note  1. — In  each  of  these  two  examples  the  probability  of  the 
event  is  the  sum  of  the  probabilities  of  the  separate  events.  It  is 
important,  however,  to  notice  that  the  probability  of  one  or  other  of 
a  series  of  events  is  the  sum  of  the  probabilities  of  the  separate  events, 
only  when  the  events  are  mutually  exclusive,  i.e.,  when  the  occurrence 
of  ar.y  one  is  incompatible  with  the  occurrence  of  any  of  the  others. 

3.  From  20  tickets  marked  with  the  first  20  numerals,  one 
is  drawn  at  random  :  find  the  chance  that  it  is  a  multiple  of 
3  or  of  7. 

The  chance  that  the  number  is  a  multiple  of  3  is  ._,';,.  and 
the  chance  that  it  is  a  multiple  of  7  is  .fv  ;  and  these  events 
are  mutually  exclusive. 

.-.     the  required  chance  =  .{'„  +  L,-0  =  -?.. 

But  if  the  question  had  been  :  find  the  chance  that  the 
number  is  a  multiple  of  3  or  of  7.  it  would  have  been 
incorrect  to  reason  as  follows  : 

Because  the  chance  that  the  number  is  a  multiple  of  8  is 
,,';,.  and  the  chance  thai  the  number  is  a  multiple  of  1  is  .,";,. 
therefore  the  chance  that  it  is  a  multiple  of  <">  <>r  1  is  ._,';,  +  ./;,, 


MUTUALLY   EXCLUSIVE   EVENTS.  191 

or  £J.  For  the  number  on  the  ticket  might  be  a  multiple 
both  of  3  and  of  4,  so  that  the  two  events  considered  arc  not 
mutually  exclusive 

Note  2. — The  distinction  between  simple  and  compound  events 
is,  in  many  cases,  purely  an  artificial  one;  it  often  amounts  to  nothing 
more  than  a  distinction  between  two  different  ways  of  viewing  the 
same  occurrence. 

4.  A  bag  contains  3  white  and  4  black  balls;  if  2  balls 
are  drawn  what  is  the  chance  that  one  is  white  and  the  other 
black? 

(1)  Regarding  the  occurrence  as  a  simple  event,  the 
chance  =  3  x  4  -=-  7C2  =  f. 

(2)  Regarding  the  occurrence  as  the  happening  of  one  or 
other  of  the  two  following  compound  events  : 

(a)  drawing  a  white  and  then  a  black  ball,  the  chance  of 
which  is  8   y   4-    4. 

7     A    "6    —     14  ' 

(b)  drawing  a  black  and  then  a  white  ball,  the  chance  of 

which  is  4        3  _    4 

7   x  T;  —  T4- 

And  since  these  events  are  mutually  exclusive,  the  required 
chance  =  T4X  +  tj  =  7- 

5.  What  is  the  chance  of  throwing  an  ace  in  the  first  only 
of  two  successive  throws  with  an  ordinary  die  ?         Ans.  ■£$. 

6.  Three  cards  are  drawn  at  random  from  an  ordinary 
pack  :  find  the  chance  that  they  will  consist  of  a  knave,  a 
queen,  and  a  king.  Ans.  g  I  !;,-,. 

7.  Find  the  chance  of  throwing  an  ace  and  only  one  in 
two  successive  trials  with  one  die.  Ans.  £#. 

8.  In  one  throw  with  a  pair  of  dice  find  the  chance  that 
there  is  neither  an  ace  nor  doublets.  Ans.  ,;. 

9.  When  6  coins  are  tossed  what  is  the  chance  that  one, 
and  one  only,  will  fall  with  the  head  up  ?  Ans.  ^. 

10.  The  odds  against  a  certain  event  are  5  to  2,  and  the 
odds  in  favor  of  another  event  independent  of  the  former 
are  6  to  5  :  find  the  chance  that  one  at  least  of  the  events 
will  happen.  •  Ans.  -ti. 


192  EXPECTATION  —  GENERAL   PROBLEM. 

231.  Expectation.  —  The  value  of  a  given  chance  of 
obtaining  a  given  .sum  of  money  is  called  the  expectation. 

Suppose  that  there  are  a  +  b  tickets  in  a  lottery  for  a 
prize  .1/;  then  since  each  ticket  is  equally  likely  to  win  the 
prize,  and  a  person  who  possesses  all  the  tickets  must  win. 
the  money  value  of  each  ticket  is  M  -=-  (a  -f  1>) ;  in  other 
words  this  would  be  a  fair  sum  to  pay  for  each  ticket; 
hence  a  person  who  held  a  tickets  might  reasonably  expect 
aM  -r-  (a  +  b)  as  the  price  to  be  paid  for  his  tickets  ;  that 
is,  he  would  estimate  this  amount  as  the  value  of  his  chance, 
or  expectation.  Therefore,  if  E  be  the  expectation,  we 
have  £  _  _  a 

a 

Thus,  the  expectation  is  the  sum  which  may  be  toon  multiplied 
by  the  chance  of  winning  it. 

1.  A  bag  contains  3  white  and  7  black  balls.  Find  the 
expectation  of  a  man  who  is  allowed  to  draw  a  ball  from 
the  bag,  and  who  is  to  receive  $1  if  he  draws  a  black  ball, 
and  $5  if  he  draws  a  white  one. 

The  chance  of  drawing  a  black  ball  is  T7^  ;  and  therefore 
the  expectation  from  drawing  a  black  ball  is  $0.70.  The 
chance  of  drawing  a  white  ball  is  ^V)  ;  and  therefore  the 
expectation  from  drawing  a  white  ball  is  $1.00.  Hence,  as 
these  events  are  exclusive,  the  whole  expectation  is  $2.20. 

2.  A  purse  contains  2  sovereigns,  3  half-crowns,  and  7 
shillings.  What  should  be  paid  for  permission  to  draw  (1) 
one  coin,  and  (2)  two  coins?     Ans.  (1)  As.  6]<1.  ;  (2)  9s.  Id. 

3.  Two  persons  toss  a  dollar  alternately  on  condition  that 
the  first  who  gets  "heads"  win  the  dollar:  find  their  ex- 
pectations. Ans.  0.66|  ;  0.3af 

232.  General  Problem. —  The  2^'obability  of  the  hap- 
pening of  an  event  in  one  trial  being  known,  required  the 
probability  of  Us  happening  once  twice,  three  times,  etc., 
exactly  in  n  trials. 

Letjj  be  tin'  probability  <>!'  the  happening  <>t'  the  event  in 
one  trial,  and  q  the  probability  of  its  failing,  so  that  q=l—j>. 


GENERAL    PROBLEM.  493 

Then  the  probability  that  in  n  trials  the  event  will  happen  in 
one  particular  tried,  and  fail  in  the  other  n  —  1  trials  is 
pq"~l  (Art.  229)  ;  and  since  there  are  n  trials,  the  proba- 
bility of  its  happening  in  some  one  of  these  and  failing  in 
the  rest  is  npqn~1.  The  probability  that  in  n  trials  the  event 
will  happen  in  two  particular  trials,  and  fail  in  the  other 
n  —  2  trials,  is  p'2q"~- ;  and  there  are  „C2  ways  in  which  the 
event  may  happen  twice  and  fail  n  —  2  times  in  n  trials  ; 
therefore  the  probability  that  it  will  happen  exactly  twice  in 
n  trials  is  „02p2g"-2.  In  general,  the  probability  that  in  n 
trials  the  event  will  happen  in  any  particular  set  of  r  trials 
and  fail  in  the  other  n  —  r  trials  is  prqn~r ;  and  as  a  set  of 
r  trials  can  be  selected  in  nCr  ways,  all  of  which  are  equally 
probable  and  are  mutually  exclusive,  the  required  probability 
of  the  event  happening  exactly  r  times  in  n  trials  is  nCrp'qn~r. 

Thus,  if  (p  +  q)n  be  expanded  by  the  binomial  theorem, 
the  successive  terms  will  be  the  probability  of  the  happening 
of  the  event  exactly  n  times,  n  —  1  times,  n  —  2  times, 
etc.,  in  n  trials. 

If  the  event  happens  n  times,  or  fails  only  once,  twice, 
.  .  .  .  (n  —  r)  times,  it  happens  r  times  or  more  ;  therefore 
the  chance  that  it  happens  at  least  r  times  in  n  trials  is 

V"    +    nC^-'q    +    nC2pn-V    +    •    •    •    +    nCn-rPrqn-r,    (1) 

or  the  sum  of  the  first  n  —  r  +  1  terms  of  the  expansion 

of  (2>  +  q)n- 

1.  In  four  throws  with  a  pair  of  dice,  what  is  the  chance 
of  throwing  doublets  twice  at  least? 

In  a  single  throw  the  chance  of  doublets  is  ^,  or  f.  ;  and 
the  chance  of  failing  to  throw  doublets  is  f ;  therefore 

P  =  h  ?  =  h  n  =  4>  r  =  2- 

The  required  event  follows  if  doublets  are  thrown  four  times, 
three  times,  or  twice  ;  hence  the  required  chance  is  the  sum 
of  the  first  three  terms  of  (1). 

Thus  the  chance  =  L(l  +  4  •  5  +  G  •  52)  =  fVx- 
64 


494  CI  SI  R  l  L   PROBLEM  —  EXAMPLES. 

2.  In  five  throws  with  .1  single  die,  what  is  the  chance 
(1)  of  throwing  exactly  throe  aces?  and  (2)  of  throwing  at 
least  three  aces? 

Herep  =  £,  q  =  f,  n  =  5,  r  =  3.     Hence 

(1)  the  chance  =  .C^tf-fl)"  =  f^|(*)8(f)3    =  *¥&■ 

(2)  the  chance  =  (i)5+  6 (*)*(§)  +  ~(£)3(t)2  =  >>¥&■ 

3.  Find  the  chance  that  a  person  with  two  dice  will  throw 

aces  at  least  four  times  in  six  trials.  Ans.  —  '*'  -'. 

(3G)6 

4.  Find  the  chance  of  throwing  an  ace  with  a  single  die 

4.1      4.  ■      •     4.  •  1  a        31031 

once  at  least  in  six  trials.  Ans.  — - — . 

66 

Notp:.  —  The  subject  of  Probability  is  so  extensive  that,  in  an 
elementary  work  on  Algebra,  it  is  impossible  to  give  more  than  a 
very  brief  sketch.  The  student  who  wishes  to  investigate  this  subject 
further,  is  referred  to  Todhunter's  Algebra,  and  Hall  and  Knight's 
Higher  Algebra;  also  to  Whitworth's  Choice  and  Chance,  an  admira- 
ble English  work;  and  when  he  becomes  acquainted  with  the  Integral 
Calculus  he  may  consult  Professor  Crofton's  article  Probability,  in 
the  Encyclopaedia  Britannica.  For  a  complete  account  of  the  origin 
and  development  of  the  subject,  see  Todhunter's  History  of  the 
Theory  of  Probability  from  the  time  of  Pascal  to  that  of  Laplace; 
in  this  work  the  reader  is  introduced  to  almost  every  process  and 
every  species  of  problem  which  the  literature  of  the  subject  can 
furnish.  For  the  practical  applications  of  the  theory  of  Probability 
to  commercial  transactions,  we  may  refer  to  the  articles,  Annuities 
and  Insurance  in  the  Encyclopaedia  Britannica. 

EXAMPLES. 

1.  Find  the  chance  of  drawing  a  red  ball  from  a  bag 
which  contains  5  white  and  7  red  balls.  Ans.  ,7,. 

2.  Two  balls  are  to  be  drawn  from  a  bag  containing  ."»  red 
and  7  white  balls:  find  the  chance  that  they  will  both  be 
white.  Ans.  ./.,. 

3.  Show  that  the  odds  are  7  to  8  against  drawing  2  red 
balls  from  a  bag  containing  '■)  red  and  2  white  balls. 


EXAMPLES.  495 

4.  A  bag  contains  5  white,  7  black,  and  4  red  balls  :  find 
the  chance  that  3  balls  drawn  at  random  are  all  white. 

Ans.  -5V 

5.  If  4  coins  are  tossed,  find  the  chance  that  there  will 
be  2  heads  and  2  tails.  Ans.  |. 

6.  If  from  a  pack  4  cards  are  drawn,  find  the  chance  that 
they  will  be  the  4  honors  of  the  same  suit.         Ans.  ^^7T5. 

7.  In  shuffling  a  pack  of  cards,  4  are  accidentally  dropped  : 
find  the  chance  that  the  missing  cards  should  be  one  from 
each  suit.  Ans.  ^VsVo- 

8.  A  has  3  shares  in  a  lottery  containing  3  prizes  and  9 
blanks;  B  has  2  shares  in  a  lottery  containing  2  prizes  and 

6  blanks  :  compare  their  chances  of  success. 

Ans.  952  to  715. 

9.  Find  the  odds  against  throwing  one  of  the  two  numbers 

7  or  11  in  a  single  throw  with  two  dice.  Ans.  7  to  2. 

10.  If  from  a  lottery  of  30  tickets  marked  1,  2,  3,  ...  . 
four  tickets  be  drawn,  find  the  chance  that  1  and  2  will  be 
among  them.  Ans.  Tf-^. 

11.  Supposing  that  it  is  8  to  7  against  a  person  who  is 
now  30  years  of  age  living  till  he  is  60,  and  2  to  1  against  a 
person  who  is  now  40  living  till  he  is  70  :  find  the  chance 
that  one  at  least  of  these  persons  will  be  alive  30  years 
hence.  Ans.  ff. 

12.  A  party  of  13  persons  take  their  seats  at  a  round 
table  ;  show  that  it  is  5  to  1  against  two  particular  persons 
sitting  together. 

13.  The  chance  that  A  can  solve  a  certain  problem  is  one- 
fourth  ;  and  the  chance  that  B  can  solve  it  is  two-thirds  : 
find  the  chance  that  the  problem  will  be  solved  if  they  both 
try.  Ans.  f . 

14.  The  odds  against  A  solving  a  certain  problem  are  4 
to  3,  and  the  odds  in  favor  of  B  solving  the  same  problem 
are  7  to  5  :  what  is  the  chance  that  the  problem  will  be  solved 
if  they  both  try?  Ans.  Jf. 

15.  A  lottery  has   1   prize  of  $50,  2  prizes  of  $5  each, 


400  EXAMPLES. 

4  prizes  of  $1  each,  and  13  blanks:  find  the  expectation  of 
the  holder  of  1  ticket.  Ans.  $3.20. 

16.  In  three  throws  with  a  pair  of  dice,  find  the  chance 
of  having  doublets  one  or  more  times.  Ans.  1  —  (|)3. 

17.  Find  the  chance  of  throwing  double  sixes  once  or 
oftener  in  3  throws  with  a  pair  of  dice.         Ans.  1  —  (ff)3. 

18.  If  on  an  average  9  ships  out  of  10  return  safe  to 
port,  find  the  chance  that  out  of  5  ships  expected  at  least  3 
will  arrive.  Ans.  $$£§§• 

19.  If  four  cards  be  drawn  from  a  pack,  find  the  chance 
that  they  will  be  marked  one,  two,  three,  four,  of  the  same 

4-  14 

suit.  Ans.  . 

52  -51-  50  -49 

20.  Two  persons,  A  and  B,  engage  in  a  game  in  which 
A's  skill  is  to  B's  as  2  to  3.  Find  the  chance  of  A's  win- 
ning at  least  2  games  out  of  5.  Ans.  . 

21.  As  chance  of  winning  a  single  game  against  B  is  § : 
find  the  chance  of  his  winning  at  least  2  games  out  of  3. 

Ans.  -j%V 

22.  What  is  the  chance  of  throwing  at  least  2  sixes  in  6 
throws  with  a  die?  Ans.  Hfti- 

23.  A  person  goes  on  throwing  a  single  die  until  it  turns 
up  ace.  What  is  the  chance  (1)  that  he  will  have  to  make 
at  least  ten  throws ;   (2)  that  he  will  have  to  make  exactly 

ten  throws?  Ans.  (1)  (f)9;  (2)  i 

24.  A  die  is  to  be  thrown  once  by  each  of  four  persons, 
A,  B,  C,  D,  in  order,  and  the  first  of  them  who  throws  an 
ace  is  to  receive  a  prize.  Find  their  respective  chances,  and 
(5)  the  chance  that  the  prize  will  not  be  won  at  all. 

Ans.  A's  chance  £,  B's  -fa,  C's  ■££$,  D's  ^sftfr,  (5)  T6^- 

25.  What  is  the  chance  that  a  person  with  two  dice  will 
throw  aces  exactly  four  times  in  six  trials?  Ans.  7., -,:,,rl'i  u- 


GENERAL   EQUATION   OF   THE  n"'   DEGREE.  497 


CHAPTER    XXIV. 

THEORY     OF     EQUATIONS. 

233.    General  Equation   of  the  nth  Degree.  —  The 

general  form  of  an  equation  of  the  nih  degree  with  one 
unknown  quantity  is 

xn  +  p.x"-1  +  p2xn-2  + +  Pn-jc  +  pH  =  0,    (1) 

where    n   is    a    positive    integer,    and    the    coefficients    pv 

Pv p„  are  either  positive  or  negative,  integral  or 

fractional.  Tlie  coefficient  of  a?  may  always  be  made  unity, 
as  above,  by  dividing  the  equation  by  the  coefficient  of  xn. 
The  term  pn  may  be  considered  as  the  coefficient  of  x°,  and 
is  called  the  absolute  term.  This  equation  is  called  the 
General  Equation  of  the  n"'  Degree,  because  it  is  the  form  to 
which  every  Algebraic  equation  can  be  reduced. 

Unless  the  contrary  is  specified,  the  n  coefficients  pv 
jD2, pn  will  always  be  supposed  rational. 

When  one  quantity  is  so  connected  with  another  that  no 
ehange  can  take  place  in  the  latter  without  producing  a 
corresponding  change  in  the  former,  the  former  is  called 
a  function  of  the  latter.  Any  Algebraic  expression  which 
contains  x  is  an  Algebraic  function  of  x,  and  may  be  denoted 
by  f(x).*  Thus,  the  first  member  of  (1)  is  a  function  of  x, 
and  may  be  denoted  by  f(x) ;  then  we  may  represent  the 
general  equation  of  the  n,h  degree  by  f(x)  =  0. 

Any  value  of  x  which,  being  substituted  for  a*,  satisfies 
the  equation,  i.e.,  makes  f(x)  equal  to  0,  is  called  a  root 
of  the  equation  (Art.  53). 

We  shall  assume  that  every  equation  of  the  form  f(x)  =  0 
has  a  root,  real  or  imaginary.     The  proof  of  this  theorem  is 

*  This  is  read  "  a  function  of  x,"  or  "  the  /  function  of  x,"  or  more  briefly,  "  th« 
/  of  x." 


498  DIVISIBILITY   OF  EQUATIONS. 

given  in  special  treatises  on  the  Theory  of  Equations;  but 
it  is  beyond  the  range  of  the  present  work. 

An  equation  of  the  third  degree  with  one  unknown  quan- 
tity is  called  a  cubic  equation. 

An  equation  of  the  fourth  degree  with  one  unknown 
quantity  is  called  a  biquadratic  equation. 

234.  Divisibility  of  Equations.  —  (1)  If  a  is  a  root 
of  the  equation  f(x)  =  0,  then  f(x)  is  exactly  divisible  by 
x  —  a. 

For,  divide  f(x)  by  x  —  a  until  a  remainder  is  obtained 
which  does  not  contain  x ;  let  Q  be  the  quotient,  and  R  the 
remainder  if  there  be  one.  Then,  since  the  dividend  is  equal 
to  the  product  of  the  divisor  and  the  quotient,  added  to  the 
remainder, 

(x  -  a)  Q  +  R  =  0. 

But  x  =  a;  therefore  (x  —  a)Q  =  0,  and  .*.  R  =  0. 
That  is,  x  —  a  divides  f(x)  without  a  remainder. 

(2)  Conversely,  if  f(x)  is  exactly  divisible  by  x  —  a,  then 
a  is  a  root  of  the  equation  f(x)  =  0. 

For,  let  Q  be  the  quotient  when  f{x)  is  divided  by  x  —  a ; 

then 

(x  -  a)Q  =  0. 

This  equation  is  satisfied  when  x  =  a,  and  therefore  a  is  a 
root  of  the  equation  f(x)  =  0. 

1.  Prove  by  this  method  that  2,  3,  and  1  are  roots  of  the 
equation  x3  -  9x2  +  26a;  -  24  =  0. 

2.  Prove  that  4  is  a  root  of  the  equation 

xs  +  a,a  _  Ha  -  24  =  0. 

3.  Prove  that  —1  is  a  root  of  the  equation 

xi  _  38a8  +  210a;2  +  538a)  +  289  =  0. 

(8)   To  find  the  remainder  when  any  rational  function,  oj 

x  is  divided  by  x  —  a,  where  a  is  any  constant. 


NUMBER   OF  ROOTS.  499 

Let  f{x)  denote  any  rational  function  of  x ;    divide  f(x) 
by  x  —  a  until  the  remainder  is  independent  of  x ;    let  Q 
denote  the  quotient,  and  R  the  remainder ;  then 
f{x)  =  (x-  a)Q  +  R. 
Since   R    does   not    involve   x   it   will   remain    unaltered 
whatever  value  we  give  to  x  ;  put  x  =  a,  then 
/(«)  *  =  0  x  Q  +  R ; 

now  Q  is  finite  for  finite  values  of  x ;  hence  R  =  /(«) . 

235.  Number  of  Roots.  —  Every  equation  of  the  n"' 
degree  has  n  roots,  and  no  more. 

We  have  seen  (Art.  140)  that  a  quadratic  equation  has 
two  roots.  In  the  same  way,  a  cubic  equation  has  three 
roots,  and  so  on. 

Denote  the  given  equation  by  f(x)  =  0,  where  (Art.  233) 

f(x)  =  xn+  pjof-1  +  p.^""2  +  •  •  •  +  Pn-i®  +  P«  =  0-    (l) 
The  equation  f{x)  =  0  has  a  root,  real  or  imaginary  ;  let 
this  be  denoted   by  a;    then,  by   (1)   of  Art.  234,  f(x)  is 
divisible  by  x  —  a,  so  that  (1)  becomes 

fix)  =  im-a)fxix)  =  0    ....  (2) 

where  f(x)  is  a  rational  function  of  (//-  —  1)  dimensions. 

Now  (2)  may  be  satisfied  by  making  either  factor  equal  to 
0;  hence  fx(x)  =  0;  and  the  equation  fix)  =  0  must  have 
a  root,  real  or  imaginary  ;  let  this  be  denoted  by  b  ;  then 
fix)  is  divisible  by  x  —  6,  so  that  (2)  becomes 

fix)  =  ix-  a)ix  -  b)f,ix)  =  0,  .     .     .   (3) 

where  fix)  is  a  rational  function  of  (n  —  2)  dimensions. 

Continuing  in  this  way,  the  equation  fix)  —  0  will  ulti- 
mately be  resolved  into  n  binomial  factors  of  the  form  x  —  a, 
x  —  6,  x  —  c,  etc.     Hence,  calling  the  last  root  I,  (1)  may 
be  written  in  the  form 
fix)  =  ix  -  a)ix  -  b)ix  -c) ix  -  I)  =  0  .   (4) 

*  /(«)  is  the  expression  obtained  by  substituting  a  for  x  in  /(x). 


500    RELATIONS  BETWEEN   COEFFICIENTS  AND  ROOTS. 

Therefore  the  equation /(a;)  =  0  has  n  roots,  since /(a;)  is 

equal  to  0  when  x  has  any  of  the  values  a,  b,  c, I. 

And  the  equation  cannot  have  more  than  n  roots,  for  if  we 
give  to  x  a  value  different  from  any  of  the  n  values  a,  b,  c, 

I,  all  the  factors  of  (4)  are  different  from  zero,  and 

therefore  f(x)  cannot  be  zero. 

In  the  above  investigation  some  of  the  quantities,  a,  b,  c, 
....  may  be  equal ;  in  this  case,  however,  we  shall  suppose 
that  the  equation  still  has  n  roots,  although  two  or  more  of 
them  are  equal.     Thus,  factoring  the  equation 

x*  -  x2  -  8x  +  12  =  0, 
it  becomes    (x  -  2)  (re  -  2)  (a:  +  3)  =  0, 
and  hence  the  three  roots  are  2,  2,  —  3  ;  and  the  equation 
has  two  equal  roots. 

It  follows  from  this  theorem  that  if  we  know  one  root  of 
an  equation,  we  may  by  division  reduce  it  to  another  equation 
of  the  next  lower  degree,  which  contains  the  remaining  roots. 
Hence,  when  all  the  roots  of  an  equation  but  two  are  known, 
it  may  be  reduced  to  a  quadratic  by  division,  and  the  remain- 
ing roots  found  by  the  rules  for  quadratics. 

1.  One  root  of  the  equation  x3  -  9a;'2  +  2Gx  —  24  =  0 
is  3  :  find  the  other  roots.  Ans.  2,  4. 

2 .  Two  roots  of  the  equ ation  x4 - 1 2a-3 + 48a;9  -  68a;  +15  =  0 
are  3  and  5  :  find  the  other  roots.  Ans.  2  ±  y3. 

3.  Three  roots  of  a;5  +  6a;4-  10a;3-  112a;2  -  207a;  -110  =  0 
are  —1,  —2,  —5  :  find  the  other  roots.  Ans.  1  ±  ^12. 

236.  Relations  between  Coefficients  and  Roots.  — 

To  investigate  the  relations  betiveen  the  coefficients  and  the  roots 
of  any  given  equation. 

If  we  form  an  equation  with  the  two  roots,  a  and  b,  we 
have  (Art.  235) 

(x  -  a)  {x  -  b)  =  0 ; 

and  performing  the  multiplication  indicated,  this  becomes 

a?  —  (a  +  b)x  +  ab  =  0. 


RELATIONS   BETWEEN   COEFFICIENTS  AND  ROOTS.    501 

Similarly,  if  we  form  an  equation  with  the  three  roots,  a, 
b,  and  c,  we  have 

x3  —  (a  +  b  +  c).r  +  (afi  +  ac  +  6c)  a;  —  a&c  =  0  ; 

and  so  on.     Comparing  this  with  the  general  form  (see  (1) 
of  Art.  233),  we  see  that 

Pj  =  —  (a  +  b  +  c) ;    p.2  =  (ab  +  ac  +  5c) ;    p3  =  —abc. 
If  we  add  another  root,  d,  we  find  the  result  to  be 
px  =  —  (o  +  6  +  c  +  d) ; 
pa  =  (aft  +  ac  +  ad  -f  6c  +  bd  +  cd) ; 
p3  =  —  (abc  +  a&d  +  acd  +  &cd) ; 
pi  =  a&cd". 

Hence,  generally,  we  have  the  following 
Rule. 

2%e  coefficient  of  the  second  term  of  the  general  equation  is 
the  sum  of  all  the  roots  with  their  signs  changed. 

The  coefficient  of  the  third  term  is  the  sum  of  the  products 
>f  all  the  roots,  taken  tioo  and  two. 

The  coefficient  of  the  fourth  term  is  the  sum  of  the  products 
of  all  the  roots,  taken  three  and  three,  with  their  signs  changed, 
etc. 

The  last  term  is  the  product  of  all  the  roots  ivith  their  signs 
changed. 

Note.  —  It  will  be  seen  that  these  relations  include  those  of 
quadratics  given  in  Art.  140.  If  the  second  term  of  an  equation  is 
wanting,  the  sum  of  the  roots  will  be  0.  The  absolute  term  of  an 
equation  is  divisible  by  every  rational  root.  If  the  absolute  term  is 
wanting,  one  root  at  least  will  be  0. 

Form  the  equation  whose  roots  are 

1.  3,  4,  -5.  Ans.  xz  -  2a;2  -  23x-  +  GO  =  0. 

2.  3,  -2,  7.  x3  -  8x2  +  x  +  42  =  0. 

3.  2,  3,  5,  -G.        aj*  -  4x3  -  29a8  +  156a;  -  180  =  0. 


502  FRACTIONAL   ROOTS — IMAGINARY  ROOTS. 

237.  Fractional  Roots.  —  An  equation  having  unity 
for  the  coefficient  of  the  first  term,  and  all  the  other  coefficients 
integers,  cannot  have  a  root  which  is  a  rational  fraction. 

If  possible,  let  -,  a  rational  fraction  in  its  lowest  terms, 
b 
be  a  root  of 

x"  +  p.af-1  +  p«x»--  + +  pn_lX  +  pn  =  0,   (1) 

where  all  the  coefficients  are  integers. 

Substituting  -  for  x  in  (1),  multiplying  by  b"'1,  and  trans- 
posing, we  have 

|-  =  ~(p1an-i  +  p.2a"~2b  +  psan~slr  + +  pj*"1). 

That  is,  we  have  a  fraction  in  its  lowest  terms  (Art.  225) 
equal  to  the  sum  of  a  number  of  integers,  which  is  impossi- 
ble.    Therefore  (1)  cannot  have  a#  root  which  is  a  rational 

fraction. 

Note.  —  This  proposition  only  proves  that  the  rational  roots  of  the 
equation  must  be  integers.  The  equation  may,  however,  have  irrOr 
tional  fractions  as  roots.    Thus,  in  the  equation  ,r3+  a;2—  11a;  +  10  =  0, 

one  of  the  roots  is  2,  and  the  other  two  are i^_.     Such  roots  as 

2 

these,  whose  values  cannot  be  exactly  expressed,  are  called  incom- 
mensurable. 


238.  Imaginary  Roots.  —  In  an  equation  with  real 
coefficients  imaginary  roots  occur  in  pairs. 

Denote  one  of  the  imaginary  roots  of  such  an  equation  by 
a  +  hi  (Art.  219)  ;  then  if  a  +  hi  be  a  root  of  the  equation, 
a  —  hi  will  also  be  a  root. 

For,  if  a  +  hi  be  substituted  for  x  in  the  equation,  the 
result  will  consist,  (l)of  real  terms  which  involve  the  odd 
and  even  powers  of  a  and  the  even  powers  of  bi,  and  (2) 
of  imaginary  terms  which  involve   the  odd  powers  of  bi. 


DESCARTES'   RULE   OF  SFGNS.  503 

Representing  the  sum  of   all  the  real  terms  by  A,  and  the 
sum  of  all  the  imaginary  terms  by  Bi,  we  have 

A  +  Bi  =  0. 

But  in  order  that  this  equation  may  be  true,  we  must  have 
A  =  0  and  B  =  0  (Art.  220). 

If  we  now  substitute  a  —  bi  for  x  in  the  given  equation, 
the  result  will  differ  from  the  preceding  only  in  the  signs 
of  the  imaginary  terms.  Hence  we  obtain  A  —  Bi.  But 
since  -4  =  0  and  B  =  0,  we  have 

A  -  Bi  =  0. 

Thus,  a  —  bi  satisfies  the  equation,  and  therefore  it  is  also 
a  root. 

In  like  manner  it  may  be  shown  that  in  an  equation  with 
rational  coefficients,  surd  roots  enter  in  pairs  ;  i.e.,  if  a  +\b 
is  a  root,  then  a  —  ^b  is  also  a  root. 

Hence,  every  equation  of  an  odd  degree  with  real  coefficients 
has  at  least  one  real  root. 

Since  the  product  of  a  pair  of  conjugate  imaginary  factors 
is  a  real  quadratic  factor  (Art.  220) ,  it  follows  that,  in  an 
equation  of  an  even  degree,  all  the  roots  may  be  imaginary. 

1 .  One  root  of  the  equation  x3  —  15a:  +  4  =  0  is  2  +  \3  : 
find  the  other  roots.  Ans.  2  —  V^,  —4. 

2.  One  root  of  xi  -  3a;2  -  42»-40  =  0  is  -^3-^-31)  : 
find  the  other  roots.  Ans.  -£(3  +  V— 31),  4,-1. 

3.  One  root  of  6a.-4  -  13a;3  -  35a;2  -  x  +  3  =  0  is  2  -  V^  : 
find  the  other  roots.  Ans.  2  +  \^3,  —  $,  — }. 

239.  Descartes' Rule  of  Signs.  —  An  equation  f(x)  =0 
cannot  have  more  positive  roots  than  there  are  variations  of 
sign  in  f(x),  or  more  negative  roots  than  there  are  variations 
of  sign  in  f(-x). 

A  variation  of  sigu  is  said  to  occur  when  the  signs  of  two 
successive  terms  of  a  series  are  unlike;  and  a  permanence, 
when  the  signs  are  alike. 


504  DESCARTES'   RULE   OF  SIGNS. 

(1)  Suppose  that  the  signs  of  the  terms  in  any  equation 
are  -f-  H 1 1 1 —  j  i'1  which  there  are  four  per- 
manences and  seven  variations.  We  shall  show  that  if  we 
introduce  a  new  positive  root  a  into  this  equation,  which 
we  do  by  multiplying  it  by  x  —  a  (Art.  235) ,  whose  signs  are 
-\ — ,  there  will  be  at  least  one  more  variation  of  sign  in 
the  product  than  in  the  original  equation.  Writing  down 
only  the  signs  of  the  terms  in  the  multiplication,  we  have 

+  +  --  + +  -  +  - 

+  - 


+  +  --  + +  -  + 


+  ±  -  T  +  -  T  T  +  -  +  -  + 

A  double  sign  is  placed  where  the  sign  of  any  term  in  the 
product  is  ambiguous.  It  will  be  seen  by  inspection  (1)  that 
each  permanence  in  the  multiplicand  is  replaced  by  an  ambi- 
guity in  the  product,  (2)  that  the  signs  before  and  after  an 
ambiguity  or  a  set  of  ambiguities  are  unlike,  and  (3)  that  a 
variation  is  introduced  at  the  end. 

Hence,  taking  the  ambiguous  sign  either  way,  the  intro- 
duction of  the  positive  root  +  a  has  not  increased  the  num- 
ber of  permanences,  while  it  has  added  at  least  one  variation.* 
Therefore  as  each  factor,  corresponding  to  a  positive  root, 
must  introduce  at  least  one  new  variation,  the  whole  number 
of  positive  roots  cannot  exceed  the  number  of  variations  of 
sign  of  f(x) . 

(2)  To  prove  the  second  part  of  the  rule,  put  —  x  for  x ; 
then  the  variations  in  f(x)  become  •permanences  in  /(—a;), 
and  the  negative  roots  of  the  equation  f(x)  =  0  are  the 
positive  roots  of  the  equation  /( —a?)  =  0  (see  Art.  242) .  But 
from  (1)  the  number  of  positive  roots  of  /(— »)  =  0  cannot 

*  If  the  original  polynomial  ends  with  a  permanence,  an  ambiguity  will  result  in 
the  final  polynomial.  Thin  ambiguity  furnishes  a  variation  either  with  the  preceding 
or  with  the  final  sign  just  added.  So  that,  taking  the  most  unfavorable  case,  there  is 
one  variation  added. 


DERIVED   FUNCTIONS.  505 

exceed  the  number  of  variations  in  f(—x).  Therefore  the 
whole  number  of  negative  roots  of  the  given  equation  f(x)  =0 
cannot  exceed  the  number  of  variations  in  f(—x)  =  0. 

The  following  results  follow  from  Descartes'  Rule  of 
Signs : 

(I)  Since  in  a  complete  equation  the  number  of  variations 
and  permanences  is  equal  to  the  degree  of  the  equation, 
therefore,  when  the  roots  are  all  real,  the  number  of  positive 
roots  is  equal  to  the  number  of  variations,  and  the  number  of 
negative  roots  is  equal  to  the  number  of  permanences. 

(II)  If  the  coefficients  of  a  complete  equation  are  all  posi- 
tive, the  equation  has  no  positive  root ;  thus,  x3  +  Gx~  +  H# 
+  6  =  0  cannot  have  a  positive  root. 

(III)  If  the  coefficients  of  a  complete  equation  are  alter- 
nately positive  and  negative,  the  equation  has  no  negative  root : 
thus  the  equation 

x*-  —  5a;3  +  9a;2  —  7x  +  2  =  0 

cannot  have  a  negative  root. 

(IV)  In  an  incomplete  equation,  imaginary  roots  may 
sometimes  be  discovered.     Thus  in  the  equation 

3a;4  +  12a;2  +  5a;  -  4  =  0. 
there  is  one  variation,  and  therefore  there  is  but  one  positive 
root. 

Again  f{—x)  =  3x*  +  12a;2  —  5a;  —  4  =  0 ;  here  there 
is  but  one  variation,  and  therefore  but  one  negative  root. 
Therefore  the  other  two  roots  must  be  imaginary. 

1.  Show  that  the  equation  2a;7  —  a;4  +  4a;3  —  5  =  0  has 
at  least  four  imaginary  roots. 

2.  Show  that  the  equation  x9  -f-  5a;8  —  a;3  +  7x  +  2  =  0 
has  at  least  four  imaginary  roots. 

240.  Derived  Functions.  —  To  find  the  value  off(x+h), 
when  f(x)  is  a  rationed  integral  function  of  x. 

Let  f(x)  represent  the  general  equation  of  the  ?tth  degree 
(Art.  233),  so  that 
f{x)  =  af  +  pxxn~l  +  iW-2  4- +  p„_,a;  +  pn  (1) 


506  DERIVED   FUNCTIONS. 

If  we  substitute  x  +  h  for  a  (1)  becomes 

f(x+h)  =  (x+h)n+p1(x+h)*~l+  .  .  +pn-l(xj-h)+p,  (2) 

Expanding  each  term  of  the  second  member  by  the  bino- 
mial theorem  and  arranging  the  result  in  ascending  powers 
of  h,  we  have  f(x  +  It)  = 

Xn  +  pi??-1   +  ptf?~ 2  +   ....+  p^jO!  +  P« 

+  fe[%aJ»-1  +  (n  -  1)^0;"--  -f-  (to  -  2)p.jXn~z  +  .  .  .] 


+  £[»(«  -  1)«"-2  +  («  -  1)("  -  2)/^"-3  +  .  .  .] 
LA 


(3) 


We  see  that  the  first  line  of  this  expansion  is  the  original 
function  f(x) ,  and  if  we  denote  the  coefficient  of  /*  in  the 

second  line  by  f'(x)  ;  the  coefficient  of  —  in  the  third  line  by 
f"(x),  etc.,  we  have 

f{x+h)  =f(x)  +  hf(x)  +£/*(»)  +  £/"(*)  +  •  •  +--/"(*)• 

[2.  [3_  [»L 

The    functions   /'(x),    /'(»),    fix) are    called 

thej^AsZ,  second,  third,  ....  derive* dr functions  *  off(x)  ;  the 
given  function /(.c)  is  sometimes  called  the  primitive  function. 

Examining  the  coefficients  of  /i,  —  ...  in  (3) ,  we  see  that 

each  derived  function  is  obtained  from  the  preceding  by  the 
following  rule  : 

Multiply  cadi  term  by  the  exponent  of  x  in  that  term,  and 
then  diminish  the  exponent  b;/  unity. 

Note. — When  the  student  becomes  acquainted  with  the  elements 
of  the  Differentia]  Calculus  he  will  see  that  each  derived  function  is 
the  differential  coefficient  with  respect  to  x  of  the  immediately  preced- 
ing derived  function,  and  that  the  above  development  of  J\.t  +  h)  is 
only  a  particular  case  of  Taylor's  Theorem. 

*  Called  also  derived  polynomials,  or  simply  derttattoot. 


EQUAL   ROOTS.  507 

Find  the  successive  derived  functions  of  the  following : 

1 .  2a4  -  x3  -  2a2  +  5x  —  1.     Ans.  1st  8a3  -  3a2  -  4a  +  5, 

2d         24a2  -  6x  -  4, 
3d  48a  -  6, 

4th  48. 

2.  a4  -  8a3  +  14a2  +  4a  -  8.      1st  4a3  -  24a2  +  28a  +  4, 

2d  12a2  -  48a  -f  28, 

3d  24a  -  48, 

4th  24. 

241.  Equal  Roots.  —  If  an  equation  has  r  roots  equal 
to  a,  then  the  first  derived  function  will  have  r  —  1  roots 
equal  to  a. 

Let  /(a)  and  /'(a)  denote  the  equation  and  its  derived 
function  respectively,  and  let  <£(a)  be  the  quotient  when /(a) 
is  divided  by  (a  —  a)r;  then /(a)  =  (a  —  a)r<ft(x). 

Substitute  a  +  h  for  a  ;  thus, 

/(a  +  h)  =  (a  —  a  +  ft)r<£(a  +  /*)• 

Expanding  both  members  and  neglecting  powers  of  h 
above  the  first,  we  have  (Art.  240) 

/(a)+A/'(a)+  .  .  =[(a-«y+r(a-a)'-1/i][0(a)+/i(/),(a;)+  .  .]• 

Equating  the  coefficients  of  h,  we  have 

/'(.a)  =  r(a  -  a)'"1^)  +  (a  -  a)r<f/(x). 

Thus  /'(a)  contains  the  factor  a  —  a  repeated  r  —  1  times; 
i.e.,  the  equation  f'(x)  =  0  has  r  —  1  roots  equal  to  a. 

Similarly,  it  may  be  shown  that  if  the  equation  /(a)  =  0 
has  s  roots  equal  to  6,  the  equation  f'{x)  =  0  has  s  —  1 
roots  equal  to  6  ;  and  so  on. 

From  the  above  proof  we  see  that  if  /(a)  contains  a  factor 
(a  —  a)r,  then  f'(x)  contains  a  factor  (a  —  a)r_1 ;  and  thus 
/(a)  and /'(a)  have  a  common  factor  (a  —  a)r_1.  Therefore 
if  /(a)  and  /'(a)  have  no  common  factor,  no  factor  in  /(a) 


508  EQUAL   ROOTS. 

will  be  repeated  ;  hence  the  equation  f(x)  =  0  has  or  has  not 
equal  roots,  according  as  /(as)  and  f'(x)  have  or  have  not  a 
common  factor  involving  x.  Hence,  to  obtain  the  equal  roots 
of  an  equation,  we  have  the  following 

Rule. 

Find  the  greatest  common  divisor  of  the  given  equation  and 
its  first  derived  function.  If  there  is  no  common  divisor  the 
equation  has  no  equal  roots.  If  there  is  a  common  divisor 
X>lace  it  equal  to  zero,  and  solve  the  resulting  equation. 

If  f(x)  be  divided  by  the  G.  C.  D.  of  f(x)  and  /'(as), 
the  depressed  equation  will  contain  the  remaining  roots  of 
f(x)  =  0. 

Note.  —Since  /"(as)  is  the  first  derived  function  of  f'(x),  f'"(x) 
the  first  derived  function  of  f"(x),  and  so  on,  we  see  that,  if  the  equa- 
tion ./'(.r)  =  0  has  r  roots  equal  to  a,  the  equation  /'(as)  =  0  must  have 
r  —  1  roots  =  a,  the  equation  /"(as)  =  0  must  have  r  —  2  roots  equal 
to  a,  and  so  on.  By  means  of  this  principle  we  may  sometimes  dis- 
cover the  equal  roots  of  an  equation  with  less  trouble  than  by  the  ride 
above  given. 

1.  Solve  the  equation  a4  —  lias8  +  44a;2  —  76.?  +  48  =  0, 
which  has  equal  roots. 

Here      /(as)  =  a;4  -  liar5  +  44a;2  -  76a;  +  48, 

f(x)  =  Ax3  -  33a;2  +  88.r  -  76  ; 

and  by  the  rule  (Art.  74)  we  find  that  the  G.  C.  D.  of  /(.»•) 
and  f'{x)  is  (as  —  2)  ;  hence  (as  —  2)a  is  a  factor  of  f(x)  ; 
and 

.-.     f(x)  =  (x-2y-(x*-7x+l2)  =  (x-2y(x-3)(x-i)  ; 
thus  the  four  roots  are  2,  2,  3,  4. 

Solve  the  following  equations  which  have  equal  roots: 

2.  Xs  -  2as2  -  15as  +  30  =  0.  Ans.  3,  3,  -  I 

3.  xA  -  9.r-  +   !./•  +12  =  0.  2,  2,  -1,  -3. 

4.  as4  -  Gx3  +  12a;2  -  10a;  +  3  =  0.  1,1,  1,  3. 


TRANSFORMATION   OF  EQUATIONS  —  DEFINITION.   509 


TRANSFORMATION    OF    EQUATIONS. 

242.  Definition.  —  An  equation  is  said  to  be  transformed 
when  it  is  changed  into  another  equation  whose  roots  bear 
some  assigned  relation  to  those  of  the  given  equation. 

Hem.  —  Various  transformations  of  this  kind  can  be  effected  with- 
out knowing  the  roots  of  the  given  equation;  and  examples  occur  in 
which  a  root  of  the  transformed  equation,  and  hence  the  correspond- 
ing root  of  the  given  equation  is  more  easily  found  than  by  solving  the 
given  equation  directly.  Such  transformations  arc  especially  useful 
in  the  solution  of  cubic  equations. 

Problem  I. —  To  transform  an  equation  into  another 
ichose  roots  are  those  of  the  given  equation  with  contrary 
signs. 

Let  f(x)  —  0  be  the  given  equation.  Put  —  y  for  .r,  so 
that  when  x  has  any  particular  value,  y  has  numerically  the 
same  value  but  with  the  contrary  sign  ;  then  the  required 
equation  is  f(—y)  =  0. 

Thus,   iff(x)=Xn  +  1)1Xn-1  +  2Wn-2+    ■    •    +Pn-lV  +  Pn=0, 

then  f(-y)  =  (-yf+Pl(-yY-'t+p2(-y)n--+  .  .  -pn-i?J+P,-0  ; 

and  since  the  even  powers  of  —y  are  positive  and  the  odd 
powers  negative,  the  above  equation  becomes 

t    ~  PlVn~l    +  Plf~-    +    ■    •    •    •     ±   Pn-\V    T   Pn- 

Hence  the  transformed  equation  may  be  obtained  from  the 
given  equation  by  changing  the  sign  of  every  alternate  term 
beginning  with  the  second. 

Note.  —  The  above  rule  assumes  that  the  given  equation  is  com- 
plete (Art.  34) ;  if  any  terms  are  wanting,  the  equation  may  be  ren- 
dered complete  by  introducing  the  missing  terms  with  zero  for  the 
coefficient  of  each  of  them. 

1.  Transform  the  equation  3a4  —  4xs  —  Ax  +  7  =  0  into 
another  whose  roots  are  numerically  the  same  with  contrary 
signs. 


+    1'n    =    0. 


510  PROBLEM   II. 

We  may  write  the  equation  thus, 

3^4  _  4aj  +  0-t,2  _  4a?  +  7  =  0, 
then  the  transformed  equation  by  the  rule  is 
3ar  +  4x3  +  Oar  +  Ax  +  7  =  0,    or    3a;4  +  4a:3  +  4a;  +7=0. 

243.  Problem -II.—  To  transform  an  equation  into  an- 
other whose  roots  are  equal  to  those  of  the  proposed  equation 
multiplied  by  a  given  factor. 

Let  m  denote  the  given  factor ;  put  y  =  mx,  so  that  when 
x  has  any  particular  value,  the  value  of  y  is  m  times  as  large  ; 

then  x  =  y~. 

m 

Substituting  this  in  the  general  equation  (Art.  233)  we  have 

Multiplying  by  m",  we  have 
yn  +  2hmyn~l  +  p{nvyn~2  +  .  .  +  pn-i>nn-hj  +  pnmn  =  0. 

Hence  we  have  the  following  rule  : 

Mxdtiply  the  second  term  by  the  given  factor,  the  third  term 
by  this  factor  squared,  and  so  on  to  the  last  term. 

The  chief  use  of  this  transformation  is  to  clear  an  equa- 
tion of  fractional  coefficients. 

1.    Remove  fractional  coefficients  from  the  equation 
2a;3  -  far  -^4-^  =  0. 

Multiplying  the  second  term  by  m,  the  third  by  //<--,  etc., 
by  the  rule,  we  get 

2a;3  -  |maja  —  Jm3*  +  ^.ms  =  0. 

By  properly  assuming  m  we  may  make  the  coefficients  of 
the  transformed  equation  all  integers. 

Thus,  if  m  =  4,  all  the  terms  become  integral,  and  on 
dividing  by  2  we  obtain 

a»  _  3aJ  -  x  +  6  =  0, 
in  which  the  roots  arc  four  limes   as  great  as   those   of   the 
given  equation. 


PROBLEMS    III.    AND    IV.  511 

If  the  roots  are  to  be  divided  by  any  number,  we  obtain 
the  transformed  equation  by  dividing  the  second  term  by  the 
given  number,  the  third  term  by  its  square,  and  so  on. 

2.  Find  the  equation  whose  roots  are  three  times  as  great 
as  those  of  as4  +  7a;2  —  4x  +  3  =  0. 

Ans.  if  +  G3//2  -  108//  +  243  =  0. 

3.  Find  the  equation  whose  roots  are  one-half  those  of 
x-  -  2x  +  3  =  0.  Ans.  y*  —  y  +  f  =  0. 

244.  Problem  III.  —  To  transform  an  equation  into 
another  whose  roots  are  the  reciprocals  of  the  roots  of  the 

given  equation. 

Let  f(x)  =  0  be  the  given  equation ;  put  y  =  -,  so  that 
when  x  has  any  particular  value,  the  value  of  y  is  the  recip- 
rocal of  that  value ;  then  x  —  -,  and  the  required  equation 

Thus,  if  /(»)  =  ; xn+plxn-1+p&n-'2+  .  .  +  pn^xx+  pn=0, 


men 


f(  I  \  =  I  +  J>L.  +  J^  +  .  .  +  &=l  +  Pn  =  0 ; 

W     y      y        y  v 

that  is,   p„yn  +  p„_,//"-1  +  P»-2?/re_2  +  •  •  +  P$  +  1  =  ° 
Hence,  to  effect  the  transformation,  reverse  the  order  of 
the  coefficients. 

Transform  the  following  equations  into  others  whose  roots 
are  the  reciprocals  of  the  roots  of  the  given  equations. 

1.  x*  +  x3  +  3x  +  2  =  0.     Ans.  2//4  +  3//3  +  y  +  1  =  0. 

2.  a:3  -  6a;2  +  11a?  —  6  =  0.     6//3  -  ll//2  +  6//  -1  =  0. 

3.  x«  -  5a,-5  -  7a;4  +  Qxs  +  13a;2  -  9a;  +  25  =  0. 
Ans.  25?/6  -  9//5  +  13//4  +  6//3  -  If  -  5//  +  1  =  0. 

245.  Problem  IV,  —  To  find  the  equation  whose  roots 
are  the  squares  of  those  of  a  proposed  equation. 

Let  f(x)  =  0  be  the  given  equation  ;  put  //  =  a*2,  so  that 
when  x  has  any  particular  value,  the  value  of  y  is  the  square 


512  PROBLEM    V. 

of  that  value ;  then  x  =  \y,  and  the  required  equation  is 

ftfy)  =  o. 

P^xample  1.    Find  the  equation  whose  roots  are  the  squares 
of  those  of  the  equation  x8  +  2h$~  +  jV  +  Pi  =  0. 
Put  x  =  \y,  and  transpose, 

Of  +Pi)^y  =  -{Piv  +p3)\ 

squaring,  (y2  +  2p$  +  p?)y  =  p?y2  +  2piP&  +  pi- 

.-.     ys  +  {2p,  -  p{2)y2  +  {p.?  -  2plPs)y  -  p2  =  0. 

2.  Find  the  equation  whose  roots  are  the  squares  of  the 
roots  of  the  equation  x~+hx—  7  =  0.      Ans.  y2— 39?/ +49  =  0. 

246.  Problem  V. —  To  transform  an  equation  into 
another  whose  roots  are  less  than  those  of  the  proposed  equa- 
tion by  a  given  quantity. 

Let  the  given  equation  be  f(x)  =  0,  or 

Poxn  +  p^f1  +  p,p?-2  +  ....+  pn_xx  +  pn  =  0,  (1) 

and  let  h  be  the  given  quantity.  Put  y  =  x  —  h,  so  that 
when  x  has  any  particular  value,  the  value  of  y  is  less  by  h; 
then  x  =  y  +  A,  and  the  required  equation  is 

Po(y  +  ^"  +  Pi(V  +  >0"_1  +  Pa(2(  +  k)"~2  +  •  •  +  ft  =  0, 
which,  when  arranged  in  descending  powers  of  y,  becomes 
by  (3)  of  Art.  240 

whose  roots  are  less  by  h  than  those  of  ( I ) . 

If  n  is  small,  the  transformation  by  this  method  is  easily 
effected;  but  for  equations  of  higher  degrees  it  is  quite 
tedious.     The  following  method  is  simpler. 

Denoting  the  coefficients  of  the  terms  of  (2)  by  qu  <].,, 
.  .  .  5„,  it  may  be  written 

Qoyn  +  7.2T'1  +  ^""2  +  •  •  •  +  rh-xV  +  %  =  °-  (8) 
in  which  the  coefficients  g„  q,,  .  .  .  q„  are  unknown,  whoso 
values  are  to  be  determined. 


HORNER'S  METHOD    OE  SYNTHETIC   DIVISION.       513 

For  y  substitute  its  value  x  —  ft,  and  (3;  becomes 
q0(x  -  h)»+  rh(.v  -  hy-1+  .  ..+  gB_1(a;  -  h)  +  qn  =  0,    (4) 
which  is  identical  with  (1),  since  the  value's  of  x  are  the 
same  in  (1)  and  (4).      Dividing  (4)  by  (x  —  h)  we  obtain 
the  quotient 

q0(x  -  h)"-1  +  q,(x  -  7i)n~a  +  •  •  •  •  Qn-i,     •    (;}) 

with  a  remainder  q„.  Similarly,  qn_x  is  the  remainder  found 
by  dividing  (5)  by  x  —  h,  and  the  quotient  arising  from  the 
division  is 

q0(x  -  ?i)n~2  +  q,{x  -  h)n-3  +  .  .  .  +  Qn-*; 

and  so  ou,  until  all  the  coefficients  of  (3)  are  obtained.  The 
last  quotient  is  q0,  and  is  obviously  equal  to  j)„.  Hence  we 
have  the  following  rule  : 

Divide  f(x)  by  x  —  h,  and  the  remainder  will  be  the  absolute 
term  of  the  transformed  equation.  Divide  the  quotient  thus 
found  by  x  —  h,  and  the  remainder  will  be  the  coefficient  of 
the  last  term  but  one  of  the  transformed  equation;  and  so  on. 

By  dividing  f(x)  by  x  +  h  we  shall  transform  f(x)  =  0 
iuto  an  equation  whose  roots  are  greater  by  h. 

247.    Horner's  Method  of  Synthetic  Division.  — 

This  is  an  abridged  method  of  dividing  by  detached  coeffi- 
cients. The  work  of  transforming  an  equation  may  be 
abridged  by  writing  down  only  the  coefficients  of  the  several 
terms,  zero  coefficients  being  used  to  represent  terms  corre- 
sponding to  powers  of  x  which  are  absent.  In  writing  down 
the  divisor,  the  sign  of  the  last  term  may  be  changed  ;  this 
enables  us  to  replace  the  process  of  subtraction  by  that  of 
addition  at  each  successive  stage  of  the  work.  The  first 
term  of  the  divisor  is  usually  omitted. 

The  following  example  is  an  illustration. 

1.  Find  the  quotient  and  remainder  when  3x7  —  a;6  +  31a4 
-J- 2  lie  +  5  is  divided  by  a;  +  2. 


514       HORNER'S    METHOD    OF  SYNTHETIC    DIVISION. 

The  work  stands  as  follows  : 

Coefficients.  3-1         0  +  31       0         0  +  21  +    5|-2 
'  _  6  +  14  -  28  -  6  +  12  -  24  +    6 

3  _  7  +  14  +    3  -  6  +  12  -    3  +  11 

Explanation.  —  In  this  work,  3  is  the  first  term  of  the  quotient ; 
_C>  is  the  product  of  —1  and  3;  the  sum  of  —1  and  —6  is  —7.  the 
second  term  of  the  quotient;  14  is  the  product  of  —2  and  —7;  the  sum 
of  14  and  0  is  14,  the  third  term  of  the  quotient,  and  so  on.  The  last 
term  11  is  the  remainder. 

Thus,  supplying  the  powers  of  x,  the  quotient  is 
3x6  —  7x5  +  14.f4  +  3x8  —  6xa  +  I2i  —  3,  and  the  remainder  is  11. 

We  shall  now  illustrate  the  use  of  Horner's  method  in  the 
following  transformation. 

2.  Find  the  equation  whose  roots  are  less  by  3  than  those 
of  the  equation  2.x4  —  x3  —  2x2  +  5x  —  1  =0. 

To  effect  this  transformation  we  divide  successively  by 
«  -  3,  (Art.  246). 

Changing  the  sign  of  —3,  the  operation  is  as  follows  : 

2  _  1  _     2  +  5  -       1  |+3 

6  +  15  +  39  +  132 

•     '    _  5  +  13  +  44  +  131.         .'.     q4  =  181. 

6  +  33  +  138  | 

11  +  46  +  182.                      .'.     qz  =  182. 

6  +  51  | 

17  +  97.  •••     qa  =     97- 

J] 

23  =  qx. 

Hence  the  required  equation  is 

2?/4  +  23//3  +  97?/  +  182y  +  131  =  0. 
NOTE.  — Horner's  method  is  chiefly  useful  in  numerical  work. 
3.   Find  the  equation  whose  roots  are  less  by  5  than  the 
roots  of  a;4  -  lx3  -  7sca  +  22a  +  24  =  0. 

Ana.  </4  +  16/  +  830s  +  152^  +  84  =  0. 


LIMITS   OF   THE  REAL   ROOTS   OF  AN  EQUATION.     515 

4.  Find  the  equation  whose  roots  are  greater  by  2  than 
the  roots  of  5a4  +  28a;3  +  51a:2  +  32a;  —  1  =  0. 

Ans.  5?/4  —  12y3  +  3y-  -f-  Ay  —  5  =  0. 

248.  Problem  VI.  —  To  remove  any  given  term  from 
an  equation. 

By  means  of  the  transformation  in  Art.  246  we  may 
remove  any  assigned  term  after  the  first  from  an  equation. 
Thus  if  the  term  to  be  removed  is  the  second  we  choose  h  so 
that  the  second  term  in  (2)  of  Art.  246  shall  vanish  ;  that  is, 

we  put  nPoh  =  0;     or    h  =  -A. 

np0 

If  the  term  to  be  removed  is  the  third,  we  put 

lh+  in  -  l)p1h  +  n(n~  1}M2  =  0, 

and  so  obtain  a  quadratic  to  find  h.      Similarly  we  may 
remove  any  other  term. 

1.  Remove  the  second  term  from  a;3  —  6x2  +  8a,*  —  2  =  0. 
Here    pn  =    1,  p\  =    -6,    n  =   3;    then    (Art.  246), 

x  =  y  +  h  =  y  +  2. 

Making  this  substitution  the  given  equation  becomes, 

(2/+2)3-6(y+2)2+8(y+2)-2  =  0;    or    y3-  4y-2  =  0. 

Remove  the  second  term  from  the  equations : 

2.  x3  -  6a;2  -  8a;  -  2  =  0.  Ans.  if  -  20ij  -  34  =  0. 

3.  a;4  +  4a;3  +  2a;2  —  4a;  -  2  =  0.       if  -  Ay-  +  1  =  0. 

LIMITS    OF    THE    REAL    ROOTS    OF    AN    EQUATION. 

249.  Definition.  —  The  Limits  of  a  root  of  an  equation 
are  any  two  numbers  between  which  that  root  lies.  When 
we  say  that  a  certain  number  is  a  superior  limit  of  the 
positive  roots  of  an  equation,  we  mean  that  no  positive  root 
can  be  numerically  greater  than  that  number  ;  and  when  we 
say  that  a  certain  number  is  an  inferior  limit  of  the  negative 
roots  we  mean  that  no  negative  root  can  be  numerically 
greater  than  that  number.     In  the  following  three  Articles 


516  SUPERIOR   LIMIT. 

we  shall  find  the  limits  of  all  the  real  roots,  and  also  discover 
to  some  extent  the  limits  between  which  the  real  roots 
separately  lie. 

250.  Superior  Limit.  —  ( 1 )  The  numerically  greatest 
negative  coefficient  increased  by  unity  is  a  superior  limit  of  the 
positive  roots  of  an  equation  which  is  in  the  simplest  form. 

Take  the  general  equation  of  the  nth  degree  f(x)  =  0 
(Art.  233).  Let  p  be  the  numerically  greatest  negative 
coefficient  which  occurs  in  f(x).  Then  if  such  a  value  be 
found  for  x  that  f(x)  is  positive  for  that  value  of  x  and  for 
all  greater  values,  that  value  is  a  superior  limit  of  the  positive 
roots  of  the  equation  f(x)  =  0.  It  is  evident  that  any 
positive  value  of  x  which  makes 

x"  —  p^x"-1  +  a,*-2  +  x"-3  -f-  .  .  .  +  x  +  1)  .  (1) 
positive,  will  make  f(x)  positive.  That  is,  f(x)  is  positive 
for  a  positive  value  of  x  if 


X" 

-  1 

X"  - 

-  p— 

X 

—  1 

I  _ 

xn  - 

-  1  . 

p 

1 

163),  and  therefore  if  xn  —  1  —  p- is  positive,  that  is, 

if  (x"  —  1)(  1 ,  _  ,  J  is  positive  ;  and  this  last  expression 

is  positive  if  x  —  1  >  p.  Thus  f(x)  is  positive  if  x  ==  or  > p+\ . 
Therefore  p  +  1,  or  any  number  greater,  when  substituted 
for  a;  in  (1),  will  make  the  first  term  numerically  greater  than 
the  sum  of  all  the  others,  and  thus  make  (1)  positive;  that 
is,  p  +  1  is  a  superior  limit  of  the  positive  roots  of  the 
equation  f{x)  =  0. 

(2)  In  an  equation  of  the  nth  degree  in  its  simplest  form  if 
p  be  the  numerical  value  of  the  greatest  negative  coefficient, 
and  x"~r  the  highest  power  of  x  which  has  a  negative  coeffi- 
cient,  1  +  y/p  IS  a  superior  limit  of  the  positive  routs. 

Let  f(x)  =  0  be  the  given  equation  ;  since  all  the  terms 
which  precede  x"~r  have  positive  coefficients,  it  is  evident 
that  any  value  of  x  which  makes 

x"  -  p(x"-r  +  .(•"-'- J  +  .  .  .  -f-  x-  +  .c  +  1) 


INFERIOR  LIMIT.  517 

positive  will  make  f(x)  positive.     That  is,  f(x)  is  positive 

for  a    positive  value  of  x  if  x"  —  p- —   is  positive, 

(Art.  1G3).     But  if  x  >  1 

jga-r  +  l   _    j  af-r+l 

x»  _  p >  xn  —  p -. 


Hence  f{x)  will  be  positive  when  this  second  member  is 
positive  ;  or  when  x"(x  —  1)  —  pxn~r+1  is  positive  ;  or  when 
xr~1(x  —  1)  —  p  is  positive;  or  when  (x  —  l)r  =  or  >  p. 
Thus,  f(x)  is  positive  if  x  =  or  >  1  +  \Jp  ;  therefore  1  +  Vi: 
is  a  superior  limit  of  the  positive  roots  of  the  equation 
/(a)  =  0. 

251-  Inferior  Limit.  —  To  find  the  inferior  limit  of  the 
negative  roots  of  an  equation,  change  the  sign  of  every 
alternate  term  beginning  with  the  second  ;  this  will  change 
the  signs  of  all  the  roots  of  the  given  equation  (Art.  242). 
Theu  the  superior  limit  of  the  positive  roots  of  this  trans- 
formed equation,  with  its  sign  changed,  will  be  the  inferior 
limit  of  the  negative  roots  of  the  given  equation. 

1.  Find  the  superior  limit  of  the  positive  roots  of 

aj6  +  8x*  -  Ux3  -  53a;2  +  56a  -  18  =  0. 

(1)  By  (1)  of  Art.  250  we  have  53  +  1  =  54,  as  a 
superior  limit. 

(2)  By  (2)  of  Art.  250  we  have  n  =  5,  n  -  r  =  3,  and 

._     r  __  2. 

.-.  1  +  V53,  or  9  in  whole  numbers  ;  so  that  9  is  a  superior 
limit. 

We  see  that  (2)  of  Art.  250  gives  us  the  smallest  superior 
limit,  and  it  is  evident  that  (2)  always  gives  a  smaller  limit 
than  (1),  except  when  r  =  1. 

Find  the  superior  limits  of  the  positive  roots  of 

2.  x4  -  hxz  +  37a:2  -  3a;  +  39  =  0.  Ans.  6. 

3.  a;4  +  11a;2  -  25a;  -  67  =  0.  6. 


518  LIMITS   OF  SEPARATE  ROOTS. 

252.  Limits  between  which  the  Roots  Separately 
Lie.  —  If  two  numbers,  when  substituted  for  the  unknown 
quantity  in  an  equation,  (jive  results  with  contrary  signs,  one 
root  at  least  must  lie  betiveen  these  numbers. 

Let  all  the  real  roots  of  the  equation  f(z)  =  0  be  denoted 
by  a,  b,  c,  .  .  .  I,  arranged  in  the  order  of  their  magnitude, 
a  being  Algebraically  the  smallest ;  and  let  A"  be  the  product 
of  the  quadratic  factors,  containing  the  imaginary  roots 
(Art.  238),  which  can  never  change  their  sign.  Then  (Art. 
235) 

f(x)  =  (a;  -  a)(x  -  b)(x  -  c)  ....(«  -  l)X  .   (1) 

Now  suppose  x  to  increase  gradually  from  a  quantity  less 
than  a  up  to  some  quantity  greater  than  /,  assuming  in  suc- 
cession every  intermediate  value.  As  long  as  x  is  less  than 
a,  each  simple  factor  x  —  a,  x  —  b,  .  .  .  of  fix)  is  negative, 
and  hence  the  product  is  positive  or  negative  according  an 
the  number  of  simple  factors  is  even  or  odd. 

When  x  becomes  a,  fix)  becomes  0.  When  x  becomes 
greater  than  a  and  less  than  b,  the  first  factor  x  —  o  becomes 
positive  while  all  the  other  factors  remain  negative.  Hence 
as  x  changes  from  a  value  less  than  a  to  a  value  greater  than 
a  and  less  than  b,  the  function  fix)  changes  its  sign.  Also 
when  x  becomes  b,  f(x)  becomes  0  ;  when  x  becomes  greater 
than  b  and  less  than  c,  the  second  factor  x  —  b  becomes 
positive  while  the  signs  of  all  the  other  factors  remain 
unchanged.  Hence  as  x  changes  from  a  value  less  than  b  to 
a  value  greater  than  b  and  less  than  c,  the  f  unction /(.e)  again 
chauges  its  sign :  and  so  on  for  each  real  root. 

When  the  numbers  which  give  results  with  contrary  signs 
differ  by  unity,  it  is  evident  that  the  integral  part  of  a  root 
has  been  found. 

Some  limits  betiveen  which  the  roots  must  lie  man  be  found 
by  the  folloioing  rules  : 

I.  If  two  numbers,  substituted  for  X  in  /(»),  <lit'<'  r<  sit/ls 
with  contrary  signs,  Hare  must  be  one  or  some  odd  number  of 

roots  between  these  numbers. 


EXAMPLES.  519 

II.  If  two  numbers,  substituted  for  x  in  /(a) ,  give  results 
with  the  same  sign,  there  must  be  either  no  root  or  an  even 
number  of  roots  between  these  numbers. 

These  two  rules  follow  from  the  above  demonstration. 

III.  If  the  coefficients  including  the  absolute  term  are  all 
positive,  the  equation  cannot  have  a  positive  root. 

For  the  sum  of  positive  quantities  cannot  be  zero. 

IV.  Every  equation  of  an  odd  degree  has  at  least  one  real 
root  ivhose  sign  is  opposite  to  that  of  its  last  term. 

For,  in  the  function  f(x)  substitute  for  x  the  values  -f-  oo, 
0,  —  co  successively;  then 

/(  +  oo)  =+.«>,     f(0)  =  pn,     /(-co)  =-00. 

Therefore  if  pn  is  positive  f(x)  =  0  has  a  root  between  0 
and  —oo,  and  if  pn  is  negative  f(x)  =  0  has  a  root  between 

0  and  +co. 

V.  Everu  equation  of  an  even  degree  having  its  last  term  neg- 
ative has  at  least  two  real  roots,  one  positive  and  one  negative. 

For  in  this  case, 

/(  +  oo)  =  +oo,     /(0)  =  pn,     /(-co)  =  +co; 

but  pn  is  negative  ;  heuce  /(a)  =  0  has  a  root  between  0  and 
+  oo,  and  a  root  between  0  and  —  oo. 

EXAMPLES. 

1.  Prove  that  4  is  a  root  of  the  equation 

xs  +  a-a  _  ux  +  56  =  0  (Art.  234). 

Use  Horner's  method. 

2.  Find  the  remainder  when  x5  —  4  a4  +  7&2  —  2x  +  47 
is  divided  by  x  —  5.     Use  Horner's  method.  Ans.  837. 

Solve  the  equations : 

3.  x*  +  2x*  —  41a2  —  42a  +  360  =  0,  two  roots  being 
5  and  — G.  Ans.  3,  5,  —4,  — G. 

4.  xA  —  IGa3  +  86a2  —  176a  +  105  =  0,  two  roots  being 

1  and  7.  Ans.  1,  3,  5,  7. 


520  EXAMPLES. 

5.  4.x3  +  16a2  -  9a  -  36  =  0,  the  sum  of  two  of  the 
roots  being  zero.  Arts,  f ,  — §,  —4. 

G.  4a3  +  20a2  —  23x  +  G  =  0,  the  sum  of  two  of  the 
roots  being  1.  Ans.  £,  £,  —6. 

7.  x3  —  7a  -f-  6  =  0,  the  sum  of  two  of  the  roots  being 
3.  Am.  1,  2,  —3. 

8.  x-4  —  2a;3  —  21a;2  +  22a;  +  40  =  0,  two  roots  being 
-4  and  -1.  Ans.  2,  5,  — 1,  -4. 

9.  Gx*  —  29a;3  +  40a;2  —  7x  —  12  =  0,  two  roots  being 
|  and  |.  Jn«.  f ,  f,  1  ±  ^2. 

Form  the  equations  whose  roots  are 

10.  2,  3,  -5.                               Ans.  a3  —  19a;  +  30  =  0. 

11.  -2,  4,  4.                                           a;3  —  6x2  +  32  =  0. 

12.  3,  -4,  2  ±  ^3.     x*  —  3a;8  -  15a;2  +  49a;  -  12  =  0. 

13.  |,  |,  ±^3.         Qx*  -  13a;3  -  12a2  +  39a  -  18  =  0. 

14.  0,  0,  2,  2,  -  3,  -  3.      a6  +  2x5-  11a4-  1 2aj*+  36a2  =  0. 

15.  2,  2,  -2,  -2,  0,  5. 

Ans.  a6  —  5a5  —  8a4  +  40a3  +  IGa2  —  80a  =  0. 

16.  ±4^3,  5  ±  2\^1. 

Ans.  a4  -  10a3  -  19a2  +  480a  -  1392  =  0. 

17.  1  ±  ^=2,  2  ±  V^3. 

Ans.  a4  -  6x3  +  18a-  -  26a  +  21  =  0. 

Solve  the  equations : 

18.  ;i,-4-10a8+4a2-  x  —  6  =  0,  one  root  being  i±Lzd*. 

2 

Ans.  3,  _|,  L±_vE*. 
'      3  2 

19.  6a4  -  13a8—  85a*  -  a-  +  8  =  0,  one  root  being  2-vfo 

Ans.  -|,  —5,  2±V3. 

20.  a4+ -la-s-f-r».c-+2r— 2  =  0,  one  root  being  —1  +yT-\. 

Ans.  — 1±V^2,  -1±V^T. 


EXAMPLES.  521 

21.  x*  +  4a;3  +  6a;2  +  4a:  +  5  =  0,  one  root  being  V^—  1. 

Ans.  ±V/:rl,  -2  ±  V-^l. 

22.  a;5  —  a;4  +  8a;2  —  9a;  —15  =  0,  two  roots  being  s/3, 
and  1  -  2V^T.  Ans.  -1,  ±V5,  1  ±  2V^1. 

23 .  Find  the  least  possible  number  of  imaginary  roots  of  the 
equations(l)a;4+3a;2+5.«-7  =  0,(2)x10-4a;c+a;4-2x--3  =  0, 
and  (3)  a;9-a;5+a;4+a;2+l  =  0. 

Ans.(l)  Two,  (2)  Four,  (3)  Six. 

24.  Find  the  fourth  derived  function  of  a;4  —  8a;3  +  14a;2. 

Ans.  24. 

25.  If  f(x)  =  a;4  -  12a:3  +  17a;2  -  9a;  +  7,  find  the  value 
of  f(x  +  3).  Ans.  x*  -  37a;2  -  123a;  -  110. 

26.  If  f(x)  =  a;4  +  10a:3  +  39a;2  +  88x-  +  113,  find  the  value 
of  f(x  -  4).  Ans.  a;4  -  6a;3  +  15a;2  +  1. 

Solve  the  following  equations  which  have  equal  roots. 

27.  a;4  -  9a;2  +  4a;  +  12  =  0.  Ans.  2,  2,-1,  -3. 

28.  a;4  +  2a;3  -  3a;2  -  4x  +  4  =  0.  -2,  -2,  1,  1. 

29.  a;5  -  13a;4  +  67a;3  -  171a;2  +  216a;  -  108  =  0. 

Ans.  3,  3,  3,  2,  2. 

30.  a;5-^-}-4a;2-3a;+2  =  0.   Ans.  -2, 1  ± \~3,  1  ± ^~3. 

31.  '8a;4  +  4a;3  -  18a:2  +  11a;  -  2  =  0.  J,  \,  \,  -2. 

32.  x6  -  2a;5  -  4a;4  +  12a;3  -  3ar  -  18a;  +  18  =  0. 

Ans.  ±V3,  ±^3,   1  ±  V^l. 

33.  Transform  (1)  a;3  -  4a;2  +  \x  -  \  =  0  and  (2) 
3a;4  —  5a;3  +  a;2  —  a;  +  1  =  0  into  equations  with  integral 
coefficients,  and  unity  for  the  coefficient  of  the  first  term. 

Ans.  (1)  y3  -  24 f  +  9y  -  24  =  0, 

(2)  f  -  bif  +     3f  -  9y  +  27  =  0. 

34.  Transform  (1)  a;3  +  3a:2  -f  x  —  $  =  0  and  (2) 
a?  _  9X2  _(_  ^x.  _  _i^  _  o  into  others  whose  roots  are  the 
reciprocals  of  the  roots  of  the  given  equations. 

Ans.  (1)  a:3-3a;2-9a;-3  =  0,  (2)  a;3-42a;2+441a;-49  =  0. 


522  EXAMPLES. 

35.  Find  the  equation  whose  roots  are  the  squares  of  the 
roots  of  a4  +  a3  +  2x2  +  a  +  1  =  0. 

Am.  f  +  3f  +  if  +  3y  +  1  =  0. 

36.  Find  the  equation  whose  roots  are  the  cubes  of  the 
roots  of  xs  +  3a2  +2  =  0.     Am.  y3  +  33i/2  -f  V2y  +  8  =  0. 

Find  the  equations  whose  roots  are  each  less  by  3  than  the 
roots  of 

37.  xa  -  27a  -  3G  =  0.  Arts.  ya  +  9y2  -  90  =  0. 

38.  a4-27a2-14a+120  =  0.        ?/4+12?/8+27?/2-68?/  =  <n1. 

39.  x5  -  4a;4  +  3a2  -  4a  +  6  =  0. 

^%s.  y5  +  ll?/4  +  42?/3  +  57?/2  -  13?/  =  60. 

40.  Increase  by  unity  the  roots  of  the  equation  a:3  —  5a* 
+  6a  -  3  =  0.  Am.  y3  -  Sy2  +  \%y  -  15  =  0. 

Remove  the  second  term  from  the  equations : 

41.  x3  —  6a2  +  10a  -3  =  0.     Am.  if  —  2y  -f  1  =  0. 

42.  a5-j-5a4+3a3+a2+a-l  =  0.       y5—ly3+12y2-ly  =  0. 

43.  a6  -  12a5  +  3a2  -  17a  +  300  =  0. 

Ans.  y«  -  GOy4  -  320if  -  lily2  -  113y  -  42  =  0. 

Find  the  superior  limits  of  the  positive  roots  of 

44.  a4  -  15a2  +  10a  +  24  =  0.  Ans.  5. 

45.  a5  +  7a4  -  12a3  -  49a2  +  52a  -  13  =  0.  8. 

46.  a4  +  11a8  -  25a2  -  67  =  0.  6. 

Find  the  inferior  limits  of  the  negative  roots  of 

47.  3a:3  +  2a2  -  11a  -  4  =  0.  Ans.  -5. 

48.  a4  -  4a3  -  19a2  +  46a  +  120  =  0.  -8. 


SOLUTION   OF  HIGHER   NUMERICAL   EQUATIONS.     523 


CHAPTER    XXV. 

SOLUTION     OF     HIGHER     NUMERICAL 
EQUATIONS. 

253.  Commensurable  Roots.  —  It  is  always  possible 
to  obtain  solutions  of  cubic  and  biquadratic  equations,  but 
their  general  solutions  are  complicated  and  only  of  limited 
application.  The  general  Algebraic  solution  of  equations  of 
a  degree  higher  than  the  fourth  has  never  been  obtained  ; 
and  Abel's  demonstration  in  1825  of  the  impossibility  of 
such  a  solution  is  generally  accepted  by  Mathematicians. 
If,  however,  the  equations  are  numerical,  that  is,  if  they 
have  numerical  coefficients,  the  value  of  any  real  root  may 
be  found  to  any  required  degree  of  accuracy. 

By  a  commensurable  root  is  meant  a  root  which  can  be 
exactly  expressed  as  an  integer  or  fraction  without  involv- 
ing irrational  quantities. 

If  the  root  cannot  be  exactly  expressed  without  involving 
irrational  quantities,  it  is  called  an  incommensurable  root. 

We  have  seen  (Art.  243)  that  any  equation  having  frac- 
tional coefficients  may  be  transformed  into  another  which 
has  all  its  coefficients  integers  and  the  coefficient  of  the  first 
term  unity  ;  and  such  an  equation  cannot  have  a  root  which 
is  a  rational  fraction  (Art.  237)  ;  thus  we  have  only  to  find 
the  integral  commensurable  roots  ;  and  every  such  root  is  a 
divisor  of  the  last  term  (Art.  23G).  Hence,  to  find  the 
commensurable  roots,  we  have  only  to  determine  cdl  the  divisors 
of  the  absolute  term,  and  learn  by  trial  which  of  them  are 
roots  of  the  equation. 

If  we  first  find  the  superior  and  inferior  limits  of  the  roots 
(Art.  249),  the  work  will  often  be  lessened,  as  no  integer 
need  be  tried  which  does  not  fall  within  these  limits.     The 


524  COMMENSURABLE  ROOTS. 

trial  may  bo  made  by  dividing  f(x)  by  x  —  a,  where  a  is  a 
supposed  root  [Art.  234  (2)]  ;  and  this  division  is  most 
easily  made  by  Horner's  method  (Art.  247).  It  is  usual  to 
omit  +1  and  —1  from  the  divisors  to  be  tried,  as  it  is 
simpler  to  test  whether  these  values  are  roots  by  substituting 
them  for  x  in  the  given  equation. 

1.    Find  the  roots  of  xs  -  3z2  -  8a  -  10  =  0. 

Here  by  Arts.  250  and  251,  1  +  10  is  a  superior  limit, 
and  —3  is  an  inferior  limit ;  hence  all  the  roots  of  the  given 
equation  lie  between  11  and  —3.  The  divisors  of  —10 
which  fall  between  these  limits  are  10,  5,  2,  1,  —1,  —2.  It 
is  found  by  substitution  that  +1  and  —1  are  not  roots.  We 
then  arrange  the  work  of  division  as  follows : 

Coefficients.         1  -  3  -  8  -   1 0  |J10 
For  a;  =10.  10+70  +620 


+   7  +G2  +610 
Hence  10  is  not  a  root.     We  then  try  5. 

1  _3  _   8  -10  [5_ 
For  a;  =  5.  5+10+10 

+  2+2        0 
Hence  5  is  a  root.     Next  try  2. 

1  _3  _   8  -10  \2_ 
For  x  =  2.  2-2-20 

-1  -10  -30. 

Hence  2  is  not  a  root.     A  trial  of  —2  also  fails. 

Thus  the  only  commensurable  root  is  5.  The  depressed 
equation  found  by  dividing  by  x  —  5  above,  is  .<•-+  2x  +  2  =  0; 
solving  this,  we  find -1  ±^-1  for  the  remaining  roots  (Art. 
235).     Hence  the  three  roots  are  5,  —  1  ±  V7— 1. 


RECIPROCAL   EQUATIONS.  525 

Find  the  roots  of  the  following  equations  : 

2.  x3  -  Ga;2  +  Ha  —  6  =  0.  Ans.  1,  2,  3. 

3.  a;3  +  3a;2  -  4<B  -  12  =  0.  2,  -2,  -3. 

4.  ^  +  jgg  _  29a;2  -  9a;  +  180  =  0.  3,  4,  -3,  -5. 

5.  a;5  +  6a;4  -  10x3  -  112a;2  -  207a;  -  110  =  0.        _ 

Ans.  -1,  -2,  -5,  1  ±  \/l2. 

254.  Reciprocal  Equations.  —  A  reciprocal  equation 
is  one  which  is  not  changed  when  the  unknown  quantity  is 
changed  into  its  reciprocal.  Hence,  if  a  be  a  root  of  such 
an  equation,  the  reciprocal  of  a  is  also  a  root. 

It  follows  from  Art.  244  that,  in  a  reciprocal  equation,  the 
coefficients  are  the  same  in  order  whether  read  forwards  or 
backwards.  For  this  reason  reciprocal  equations  are  also 
called  recurring  equations  from  the  forms  of  their  coefficients. 
Thus  the  following  are  examples  of  reciprocal  or  recurring 
equations:  ^  _  1Qa,8  +  26>T;2  _  Wx  +  1  =  0. 

x5  -  11a;4  +  17a8  +  17a;2  -  11a;  +  1  =  0. 
axe  —  bx5   +     ex4  —     cx~  +    bx  —  a  —  0. 

Note.  —  It  is  evident  from  the  second  definition  of  a  reciprocal 
equation  that,  when  the  degree  is  even  and  the  equal  coefficients  have 
unlike  signs,  the  middle  term  is  wanting. 

Theorem  I.  —  A  reciprocal  equation  of  an  odd  degree  has 
a  root  —1  token  the  equal  coefficients  have  the  same  sign, 
and  + 1  when  they  have  opposite  signs. 

Let  aa?  +  bx"-1  +  cxn~2  +  .  .  .  ±  car  ±  6a;  ±  a  =  0  (1) 
be  a  reciprocal  equation  of  an  odd  degree.  The  number  of 
terms  is  even,  and  the  equation  may  be  written  in  the  form, 
a(xn  ±  1)  +  6(af-2  ±  l)a  +  c(xn~4  ±  l)x2+.  .  .  =  0,  (2) 
in  which  each  exponent  of  x  in  the  parentheses  is  odd. 
When  we  take  the  upper  signs  the  equation  is  divisible  by 
x  -f  1  ;  hence  —  1  is  a  root :  when  we  take  the  lower  signs 
the  equation  is  divisible  by  x  —  1  ;  hence  +1  is  a  root. 
That  is,  —1  or  +1  is  a  root  according  as  the  equal  coelli- 
cients  have  the  same  or  opposite  signs. 


526  RECIPROCAL    EQUATIONS. 

Theorem  II.  —  A  reciprocal  equation  of  an  even  degree  has 

a  root  +1  and  a  root  —  1  when  the  equal  coefficients  have 
opposite  signs. 

Taking  the  lower  signs  of  (2),  and  regarding  n  as  even, 
we  have 
a(xn  -  1)  +  b{x"-'2  -  \)x  +  c(.-e"-4  -  l)x2  +  .  .  .  =  0,   (3) 

which  is  an  equation  of  the  specified  form.  As  each  expo- 
nent of  a-  in  the  parentheses  is  even,  (3)  is  divisible  by 
x2  —  1  ;  hence  + 1  and  —  1  are  roots. 

Therefore  by  division  an  equation  of  this  form  may  be 
depressed  two  degrees,  and  become  a  reciprocal  equation  of 
an  even  degree,  with  the  equal  coefficients  having  like  signs 
(Art.  51). 

Theorem  III.  —  A  reciprocal  equation  of  an  even  degree 
ivith  its  last  term  positive  can  be  reduced  to  an  equation  of 
half  that  degree. 

Let  the  equation  be 
ax2"+  bx*n-1+  cx2n~2+  .  .  .  kx"+  .  .  .+  cx2+  bx  -fa  =  0 ;  (4) 
dividing  by  xn  and  collecting  the  terms,  we  have 

Put  x  +  -  =  z  ;     then  x2  +  -,  =  z2  —  2  ; 
a;  x2 

*  +  b  =  {x  +  1if-i*+l)  =  *-s*-> 

xi +  h  =  (*" +  .^)2  ~ 2  =  zi  ~ 4z"' +  2 ; 

and  generally,  x"  -\ =  zn  —  nz"~2  +  .  .  .  . 

xn 

Hence,  each  of  the  binomials  may  lie  expressed  in  terms  of 
z,  and  by  substituting  in  (5)  we  have  an  equation  in  z 
of  the  ath  degree,  i.e.,  of  half  the  degree  of  (4),  the  given 
equation. 


BINOMIAL   EQUATIONS.  527 

1.  Given  2xG  -fa;5-  13a;4  +  13a;2  -  x  —  2  =  0. 

Here  +1  and  —  1  are  roots  by  inspection,  also  see  Theorem 
II.     Dividing  the  first  member  by  x'1  —  1,  we  obtain 

2a:4  +  a;3  -  11a;2  +  x  +  2  =  0. 

Divide  by  2ar,  put  »  +  -  =  «,  and  we  have 

x 

,  +  §-»,         .-.    ,=§or-3. 

1        5 
Hence  a;  -| —  =  -  or  —3  ; 

x       2 

.-.     a?  =  2  or  |  or  i(-3  ±  V^5). 
Solve  the  following  equations : 

2.  a;4-  5a;3  +  6a;2-  5a;  +  1  =  0.     Am.  2  ±  V&, 


±\/^S 


3.   aj6_a8  +  aJ*-aja  +  oj-l  =  0.      ±1,  ±^1,1±*     3- 

255.  Binomial  Equations.  —  The  general  form  of  a 
binomial  equation  is  xn  ±  k  =  0,  where  &  is  a  known  quan- 
tity. 

The  roots  of  this  equation  are  all  different,  because  xn  ±  k 
and  the  first  derived  function  nx"-1  evidently  have  no  common  . 
divisor.     See  Art.  241. 

If  xn  —  k  —  0  we  have  x  =  nyjk ;  i.e.,  x  is  equal  to  an  nih 
root  of  k.  But  xn  —  k  =  0  has  n  roots  (Art.  235) .  There- 
fore any  Algebraic  quantity  *  has  n  different  nth  roots. 

Let  a  denote  one  of  the  wth  roots  of  k,  so  that  k  =  a". 
Assume  x  =  az ;  then  a;n  —  fc  =_  0  becomes  anzn  —  a"  =  0  ; 
or  zn  —  1  =  0.  Hence  z  =  y/l,  i-e->  2  is  eclual  to  an  w'h 
root  of  unity.     And  since 

x  =  az  =  a  y/l  =  Y&  from  above,  we  have  ny'&  =  o  yl. 

*  By  an  Algebraic  quantity  here  is  meant  either  a  real  or  an  imaginary 
quantity. 


528  BINOMIAL   EQUATIONS. 

Thus  all  the  na  roots  of  any  Algebraic  quantity  may  be 
found  by  multiplying  any  one  of  them  in  succession  by  the 
values  of  the  n"1  roots  of  unity. 

(1)  If  n  is  odd,  the  equation  af  ±  1  =  0  has  a  root  —1 
or  +1  according  as  we  take  the  upper  or  lower  sigu, 
Art.  254,  Theo.  I.  If  xn  ±  1  be  divided  by  x  ±  1  it  will  be 
depressed  one  degree,  and  the  resulting  reciprocal  equatiou 
will  contain  the  remaining  n  —  \  roots. 

(2)  If  n  is  even,  the  equation  af  —  1  =  0  has  two  real 
roots  +1  and  -1,  Art.  254  Theo.  II.  If  xn  —  1  be  divided 
by  x2  —  1,  it  will  be  depressed  two  degrees,  and  the  resulting 
reciprocal  equation  will  contain  the  remaining  n  —  2  roots. 

(3)  If  n  is  even,  the  equation  xn  +  1  =  0  has  no  real 
root,  since  y/  —  1  is  then  impossible.  Hence  all  the  roots  of 
this  equation  are  imaginary. 

The  following  are  some  of  the  cases  of  binomial  equations 
which  can  be  solved  by  methods  already  given.  For  the 
general  case,  De  Moivre's  theorem  in  Trigonometry  must  be 
employed. 

t.   Solve  xb  -  1  =  0. 

Since  x5  -  1  =  (x  -  1)  (x*  +  a8  +  ar»  +  x  +  1), 
we  have  x  —  1  =  0 ;  or  else  x*  +  xs  +  x2  +  a;  +  1  =  0. 

Therefore  x  =  1  ;    and  dividing  the  latter  equation  by  xr 

and  putting  x  -\ —  =  z  we  have 
x 

*•  +  *  =  !;  ...     «  =  =if£ 

Replacing  the  value  of  z  and  solving  for  x,  we  have 

x  =  -*  +  V^  ±  jtf-10  _  2V5, 

4 

or       x  =  -1  ~  V^  ±  ^-10  +  2^5,     or    x  =  1, 
for  the  5  roots. 


CARDAN'S  SOLUTION   OF  A   CUBIC  EQUATION.       529 

2.  Solve  a;4  -  1  =  0.  Am.  1,-1,  /— h  -y7-^- 

3.  Solve  a;4  +  1  =  0.  ±1  *V-1. 

y/2 
256.  Cardan's  Solution  of    a  Cubic    Equation.— 

The  general  form  of  a  cubic  equation  is 

x3  -\-  ax2  -f  bx  +  c  =  0 ; 
but  as  we  can  always  remove  the  second  term  from  an  equa- 
tion (Art.  248),  we  shall  suppose  the  cubic  reduced  to  the 
simple  form, 

sb8  +  px  +  q  =  0 (1) 

Put  x  =  y  -\-  z;  then  substituting  in  (1)  we  have 
y3  +  z3  +  {oyz  +  p)  (y  +  z)  +  q  =  0. 

Now  since  y  and  z  are  any  two  quantities  subject  only  to 
the  condition  that  their  sum  is  equal  to  one  of  the  roots  of 
the  given  equation,  we  are  at  liberty  to  suppose  further  that 
they  satisfy  the  equation  3yz  +  p  =  0.     Thus  we  obtain 
ys  +  zs  +  q  =  0 .  #     m  (2)  3yz=  -p  .     .   (3) 

Substituting  in  (2)  the  value  of  z  in  terms  of  y  from  (3), 
we  obtain  the  quadratic 


y6  +  qif  = 

27 

Solving,  we  have  y3  =  —  -  +  \ 

A2  +  il8-  .-. 

V  4        27- 

•     -(4) 

— -5-i 

Since  a;  =  y  +  2,  we  have 

/l  +  27'     '     * 

•     -(5) 

-H+\/M4i-^ 


(6) 


Thus  the  expression  for  x  is  the  sum  of  two  cube  roots, 
and  as  every  quantity  has  three  cube  roots,  it  would  appear 
that  by  combining  each  of  the  three  cube  roots  in  the  value 
of  y,  with  each  of  the  three  cube  roots  in  the  value  of  z, 


530       CARDAN'S  SOLUTION   OF  A   CUBIC   EQUATION. 

we  should  obtain  on  the  whole  nine  values  of  x.  But  a 
cubic  equation  can  have  only  three  roots  (Art.  235),  so 
that  we  are  led  to  conclude  that  only  three  values  will  be 
admissible  for  x,  which  may  also  be  shown  as  follows : 

From  (3)  we  see  that  the  cube  roots  are  to  be  taken  in 
pairs  so  that  the  product  of  each  pair  is  rational.  Hence  if 
y,  z  denote  the  values  of  any  pair  of  cube  roots  which  fulfil 
this  condition,  and  «,  a2  the  imaginary  cube  roots  of  unity, 
then  the  only  admissible  pairs  will  be  y  and  z,  ay  and  a'2z, 
a-y  and  az,  since  the  product  of  each  of  these  three  pairs 
and  of  no  other  pair  is  rational.     Hence  the  roots  of  (1)  are 

y  +  z,  ay  +  a2z,  a2y  +  az. 

When  the  real  root  has  been  found  from  (6),  instead  of 
expressing  the  other  two  roots  by  the  same  method,  it  is 
preferable  to  divide  ( 1 )  by  x  minus  the  root  and  thus  depress 
the  equation  to  a  quadratic  (Art.  235). 

Note.  —  The  above  solution  is  commonly  known  by  the  name  of 
Cardan's  Solution,  because  it  was  first  published  by  him  in  1545. 
Cardan,  however,  was  not  the  inventor;  he  obtained  it  from  Nicholas 
Tartalea.  The  solution  ought  to  be  attributed  to  Nicholas  Tartalea 
and  Scipio  Ferreus,  who  seem  to  have  discovered  it  about  the  same 
time,  about  1505,  and  independently  of  each  other.  (See  historical 
note  in  Burnside  and  Panton's  Theory  of  Equations). 

1.    Solve  the  equation  x3  —  15cc  =  126. 
Put  x  =  y  +  z,  then  y3  +  z3  +  (3?/z  -  lo)(y  +  z)  =  126  ; 
put  3yz  -  15  =  0,  then  y3  +  z3  =  126. 

Substituting  for  z  its  value,  we  have  the  quadratic 

y*  -  126y3  +  125  =  0 

.-.     y3=  125,  z3=  1;  .-.    y  =  5,  2  =  1. 

Thus  x  =  y  +  z  =  5  +  1  =  6. 

Dividing  the  given  equation  by  x  —  6  we  get 
x2  +  6x  +  21  =  0, 
the  roots,  of  which  are  —3  ±  2v— 3. 
Ik' nee  the  three  roots  are  6,  —3  +  2>J  —  o,  —3  —  2^—3. 


CARDAN'S  SOLUTION   OF  A   CUBIC  EQUATION.       531 

If  L  _|-  3L  is  negative  the  expressions  for  y3  and  z3  are 

4        27 

imaginary,  but  when  this  is  the  case  it  may  be  shown  that 
the  roots  of  the  cubic  are  all  real  and  unequal.  As  there  is 
no  general  Arithmetic  or  Algebraic  method  of  finding  the 
exact  value  of  the  cube  root  of  imaginary  quantities,  Car- 
dan's solution  of  a  cubic  equation  is  of  little  practical  use 
when  the  roots  of  the  cubic  are  all  real  and  unequal. 

This  case  is  sometimes  called  the  Irreducible  Case  of  Car- 
dan's solution.  The  process  by  which  we  may  obtain  the 
roots  of  a  cubic  equation,  in  the  irreducible  case,  is  given 
in  works  on  Trigonometry.  This  is  a  matter  however  of  very 
little  practical  value. 

2.  Solve  x3  +  3a;2  +  9x  -  13  =  0. 

To  remove  the  second  term,  we  have  h  =  —  1,  (Art.  248). 
.-.  x  =  y  —  1  (Art.  246).  Substituting  this  value  of  x 
and  reducing,  the  transformed  equation  becomes 

y3  +  6y  =  20. 
Here  p  —  6,  q  =  —  20 ;  hence,  substituting  in  (6)  we  get 
y  =  (10  +  V^108)*  +  (10  -  VT08)*  =  2.732  -  .732  =  2. 
By  division,  we  find  for  the  depressed  equation 
y*  +  2y  +  10  =  0 ; 
therefore  the  other  two  values  of  y  are  —  1  ±  V  — 9  ;  and  since 
x  =  y  —  1,  the  corresponding  values  of  a;  are  1,  —  2+3V  — 1, 
-2  -  SV^. 

3.  Solve  x3  -  9x  +  28  =  0.  Ans.  4,  2  ±  \P^d. 

4.  Solve  x3  -  18.x  =  35.  5,  ~°  *         3- 

5.  Solve  xs  +  72x  -  1720  =  0.  10,  -5  ±  7V^3. 

Rem.  —  It  is  not  thought  worth  while  to  introduce  into  an  element- 
ary treatise  like  this  any  of  the  solutions  of  biquadratic  equations 
which  have  been  obtained  by  Ferrari,  Descartes,  Waring,  .Simpson, 


532  INCOMMENSURABLE  ROOTS. 

Euler,  and  others.  In  each  of  these  methods  we  have  first  to  solve  an 
auxiliary  cubic  equation ;  and  as  in  the,  case  of  the  cubic,  the  general 
solution  is  not  adapted  for  writing  down  the  solution  of  a  given 
numerical  equation.  Practically  the  methods  are  worth  but  little,  as 
the  value  of  any  real  root  in  a  numerical  equation  may  be  found  to 
any  required  degree  of  accuracy  by  Horner's  method  of  approximation. 

257.  Incommensurable  Roots.  —  If  a  numerical  equa- 
tion is  found  to  contain  no  commensurable  roots,  or,  if  after 
the  commensurable  roots  are  removed  the  depressed  equation 
is  still  of  a  higher  degree  than  the  second,  the  incommen- 
surable roots  must  then  be  found.  The  first  operation  is  to 
find  the  integral  parts  of  these  roots.  There  are  several  ways 
of  doing  this,  the  most  celebrated  of  which  is  by  Sturm's 
Theorem;  but  all  of  them  are  so  laborious  in  practice,  that, 
in  ordinary  cases,  it  is  generally  easiest  to  proceed  by  trial, 
substituting  entire  numbers  for  x  in  the  equation,  until  two 
consecutive  numbers  are  found  between  which  one  or  more 
roots  must  lie  (Art.  252, 1.).  The  decimal  parts  of  the  roots 
must  then  be  found  by  methods  of  approximation.  Of  these 
the  most  convenient  is  Horner's,  by  which  we  can  always 
determine  the  numerical  values  of  the  real  roots  to  any 
required  degree  of  accuracy. 

258.  Horner's  Method  of  Approximation.  —  This 
method  depends  on  the  successive  transformation  of  the 
given  equation,  so  as  to  diminish  its  roots  at  each  transfor- 
mation (Art.  24G). 

Leta?+p1rft-1+ptffl-*+.  .  .  .  +  p«-i&  +  i>«  =  0  .   (1) 

be  the  given  equation,  one  of  whose  roots  is  required  :  and 
suppose  we  have  found  a  to  be  the  integral  part  of  this  root 
so  that  the  required  root  lies  between  a  and  a  +  1;  and  lei 
a',  <i",  a'",  ...  be  the  decimal  digits  taken  in  order,  so  that 
x  =  a  +  a'  +  a"  +  .  .  .  Transform  this  equation  into  another 
whose  roots  are  less  by  a  (Art.  24G),  and  let 

?/»   +   gia?-l   +   q^"-  +  .   .   .   +  qn-,X   +  <J„   =   0    .    (2) 


HORNER'S   METHOD   OF  APPROXIMATION.  533 

be  the  transformed  equation  ;  then  one  root  of  this  equation 
is  between  0  and  1 ,  and  by  trial  we  may  easily  find  its  first 
decimal  digit  a'.  Transform  (2)  into  another  equation  whose 
roots  are  less  by  a',  and  let 

zn   +   y^n-X   +   ^n-2   +   #    <    ,    +   r%_#   +  f,  B   0    ,     (3) 

be  the  transformed  equation.  Then  one  root  of  this  equation 
is  between  .00  and  .1  ;  i.e.,  it  is  either  .00,  .01,  .02,  ...  or 
.09,  and  its  first  digit  may  easily  be  found.  Transform  (3) 
into  another  equation  whose  roots  are  less  by  a"  ;  and  so  on. 
As  «',  a",  etc.  are  fractions,  their  higher  powers  are  com- 
paratively small,  and  therefore  approximate  values  of  a',  a", 
.  .  .  may  generally  be  found  by  considering  only  the  last 
two  terms  of  (2)  and  (3)  ;  thus 

a'=  -qn  -=-  ?„_!,  nearly;  and  a"=-r„-^r„_i,  nearly.   (4) 

Having  found  a,  a' ,  a",  we  next  transform  (3)  into 
another  equation,  and  from  the  last  two  terms  of  the  result- 
ing equation,  find  a'",  the  next  decimal  figure,  and  so  on. 
The  absolute  term  of  the  equation  is  sometimes  called  the 
dividend,  and  the  coefficient  of  the  first  power  of  the  unknown 
quantity,  the  trial  divisor.  The  approximate  values  of 
the  succeeding  decimals  a',  a",  .  .  .  increase  in  accuracy  the 
smaller  they  become.  In  general,  after  three  or  four  decimal 
figures  have  been  found,  the  next  three  or  four  figures  may 
be  obtained  accurately  by  division. 

Hence,  to  find  a  positive  incommensurable  root  of  a 
numerical  equation,  we  have  the  following 

Rule. 

Find  by  trial  the  integral  part  of  the  root,  and  transform 
the  given  equation  into  another  whose  roots  are  less  than  those 
of  the  given  equation  by  this  integral  part. 


534  HORNER'S   METHOD   OF  APPROXIMATION. 

Divide  the  absolute  term  of  the  transformed  equation  by  the 
coefficient  of  the  first  power  of  x  for  the  first  decimal  figure 
of  the  root,  and  transform  this  last  equation  into  another 
tvhose  roots  are  less  than  those  of  the  second  equation  by  the 
figure  of  the  root  last  found. 

Divide  as  before  for  the  next  figure  of  the  root,  and  con- 
tinue this  process  until  the  root  is  found  to  the  required  degree 
of  accuracy. 

Note  1. — To  find  a  negative  root  of  the  given  equation,  change 
the  signs  of  the  alternate  terms  beginning  with  the  second,  and  find 
the  corresponding  positive  root  of  the  transformed  equation.  This 
by  a  change  of  sign  will  be  the  required  negative  root  of  the  given 
equation.  -  (Art.  242. ) 

Note  2.  —  It  may  happen  that,  in  obtaining  the  first  or  second 
decimal  figure  from  the  last  two  terms  of  the  equation,  we  get  a  result 
too  large.  In  such  case  the  error  will  always  be  made  manifest  by 
observing  the  signs  of  the  last  two  terms  in  the  next  transformed 
equation ;  for,  as  a',  o",  .  .  .  are  to  be  positive  numbers,  the  las!  i  \\  o 
terms  of  the  transformed  equation  must  have  opposite  siyns. 

1 .    Let  it  be  required  to  find  a  positive  root  of  the  equa- 
tion x3  -  Sx2  -  2x  4-  5  =  0. 
By  trial  we  find  that 

/(I)  =  1, 

/(2)  =  -3, 

/(3)  =  -1, 

/(4)  =  13; 

therefore  (Art.  252)  one  root  of  this  equation  lies  between 
1  and  2,  and  one  root  lies  between  3  and  4  ;  we  shall  find 
the  hitter  root.  Since  3  is  the  integral  part  of  this  root,  we 
first  transform  the  equation  into  another  whose  roots  are  less 
by  3.  The  successive  transformations  are  usually  written  in 
connection,  the  divisions  being  made  by  Horner's  Method  of 
Synthetic  Division  (Art.  247).  The  following  is  the  solution 
for  approximating  to  this  root  as  far  :is  three  places  of 
decimals,  the  coefficients  of  the  different  transformed  equa- 


HORNER'S  METHOD   OF  APPROXIMATION.  535 

tiojis  being  indicated  by  the  figures  (1),  (2),  (3),  .  .  .     The 
whole  operation  is  usually  exhibited  thus : 


-3 

-2     +5  13.128 

+3 
0 

0      -6 
-2      -1(1) 

3 
3 

9        .761 
7«     -.239<2> 

3 

6d) 

.61      .167128 
7.61    -.071872(3> 

.1 
6.1 

.62      .068273152 
8.23(2)   -.003598848^ 

.1 
6.2 

.1264 
8.3564 

.1 

.1268 

6.3<2> 

8.4832<3> 

.02 
6.32 

.050944 
8.534144 

.02 

.051008 

6.34 

8.585152(4) 

.02 
6T36(3) 

.008 
6.368 

.008 

6.376 

.008 

6.384(4> 

Here  the  figure  (1)  shows  where  the  first  transformation 
ends,  and  the  figure  (2)  shows  where  the  second  transform- 
ation ends,  and  so  on.  By  thus  marking  the  coefficients  in 
each  transformed  equation,  it  is  not  necessary  to  rewrite  it.* 

*  The  successive  transformed  equations,  (1),  (2),  (3) written  out  would  be 

(1)  y3+6y2+7(/-l=0,  (2)  8S+6.3s8+8.23e-.239=0,  (3)  m;3+6.36hP+8.4S32!C-.071S72=0  .  .. 


536  NEWTON'S  METHOD   OF  APPROXIMATION. 

To  find  the  second  figure  of  the  root  we  have  —  (  — 1)-=-7, 
so  that  .1  is  the  nearest  number  to  be  tried.  To  find  the  third 
figure  of  the  root  we  have  -(-.239)  -~  8.23,  so  that  .02  is 
the  nearest  number  to  be  tried.  To  find  the  fourth  figure 
of  the  root  we  have  -(-.071872)  --  8.4832,  so  that  .008  is 
the  nearest  number  to  be  tried.  In  each  of  these  cases  the 
number  tried  is  found  to  be  correct. 

After  obtaining  three  decimal  figures  of  the  root  the  next 
three  may  be  obtained  by  dividing  the  absolute  term  by  the 
coefficient  of  x. 

If  the  coefficient  of  x  in  any  of  the  transformed  equations 
is  zero,  the  next  figure  of  the  root  in  this  case  may  be  found 
by  dividing  the  absolute  term  by  the  coefficient  of  or,  and 
extracting  the  square  root  of  the  quotient. 

Eem. — Various  suggestions  have  been  offered  with  the  view  of 
shortening  the  work  in  the  use  of  Horner's  method.  With  respect 
to  such  suggestions,  it  may  be  well  to  quote  the  following  which 
occurs  in  connection  with  one  of  them.  'But  considering  that  the 
process  is  one  which  no  person  will  very  often  perform,  we  doubt 
whether  to  recommend  even  this  abridgment.  All  such  simplifica- 
tions tend  to  make  the  computer  lose  sight  of  the  uniformity  of 
method  which  runs  through  the  whole;  and  we  have  always  found 
them,  in  rules  which  occur  only  now  and  then,  to  afford  greater 
assistance  mforgetting  the  method  than  in  abbreviating  it."  Penny 
Cyclopaedia,  article  Involution. 

Find  a  root  in  each  of  the  following  equations : 

2.  x-3  +  4a;2  -  5a;  -  20  =  0.  Ans.  2.23608. 

3.  a;3  +  4a;2  -  9a;  -  57.623625  =  0.  3.45. 

4.  x*  _  4X*  _  8a;  +  27  =  0.  3.6796. 

259.   Newton's  Method  of  Approximation.  — Find 

by  trial  two  numbers  a  and  b,  one  less  and  the  other  greater 
than  a  root  of  the  equation  f(x)  =  0  and  very  nearly  equal 
to  it;  then  at  least  one  real  root  lies  between  these  numbers 
(Art.  252),  and  suppose  a  to  be  nearer  the  root  than  b  ;  let 
a  +  h  denote  the  exact  value  of   the  root,  so  that  h  is  a 


APPROXIMATION  BY  DOUBLE  POSITION.  537 

small  fraction  which  is  to  be  determined.  Substituting  a+h 
for  cc  in  the  given  equation,  we  have,  by  Art.  240, 

/(«  +  h)  =  /(«)  +  kf(a)  +  ~f\a)  + =  0. 

LL 

Now  since  h  is  supposed  to  be  a  small  fraction,  7t2,  7is,  .  .  . 
will  be  small  compared  with  h  ;  if  we  neglect  the  squares 
and  higher  powers  of  h  in  the  above  equation,  we  shall  have 
approximately 

/*=-/(a)-/». 

Applying  this  approximate  value  of  h  to  the  assumed  root 

a  gives  us 

«-/(«)-/», 

as  a  new  approximation  to  the  roots  of  the  given  equation. 
Denote  this  new  approximation  by  av  and  then  proceeding 
as  before  we  obtain 

«i  -  /(«i)  +  /  («i)> 

as  a  new  approximation ;  and  so  on.  (See  Todhunter's 
Theory  of  Equations) . 

The  following  example  of  Newton's  method  was  selected 
by  Newton  himself :  Find  the  value  of  x  in  the  equation, 

a;3  _  2x  —  5  =  0.  Ans.  2.09455148  nearly. 

260.  Approximation  by  Double  Position.  —  Find 
by  trial  two  numbers  a  and  6,  one  less  and  the  other  greater 
than  a  root  of  the  equation  f(x)  =  0,  and  very  nearly  equal 
to  it ;  then  at  least  one  root  lies  between  these  numbers 
(Art.  252),  and  suppose  a  to  be  nearer  the  root  than  b.  Let 
A  and  B  denote  the  results  when  a  and  b  are  separately 
substituted  for  x  in  f(x) ;  and  assume  that  the  results,  which 
may  also  be  considered  as  the  errors  of  the  results,  are  pro- 
portional to  the  errors  of  the  assumed  numbers.  Although 
this  assumption  is  not  strictly  correct,  yet  if  the  numbers  a 
and  b  are  nearly  equal  to  the  root,  the  error  of  the  assump- 


538  APPROXIMATION  BY  DOUBLE  POSITION. 

tion  is  small,  and  becomes   less   and   less   the    further   the 
approximation  is  continued.     Thus  we  have 

A  :  B  : :  x  —  a :  x  —  b. 
.'.     A  —  B:b  —  a  ::  A:x  —  a. 
Hence  we  have  approximately  the  following 
Rule.  — As  the  difference  of  the  errors  is  to  the  difference 
of  the  two  assumed  numbers,  so  is  either  error  to  the  correc- 
tion of  the  corresponding  number. 

This  correction  is  to  be  added  to  the  corresponding  num- 
ber when  it  is  too  small,  and  subtracted  when  it  is  too  great, 
and  the  result  will  be  a  nearer  approximation  to  the  true 
root.  This  result  and  another  assumed  number  such  that 
the  root  may  lie  between  them,  may  now  be  used,  for  obtain- 
ing a  second  approximate  value  ;  and  so  on. 

It  is  generally  best  to  begin  with  two  numbers  which  differ 
from  each  other  only  by  unit}' ;  and  it  is  also  best  to  use  the 
smaller  error. 

This  method  of  approximation  is  applicable  whether  the 
equation  is  fractional,  radical,  or  exponential ;  it  may  there- 
fore be  applied  to  many  equations  which  cannot  be  solved  by 
either  of  the  preceding  methods. 

1.  Find  a  root  of  the  equation  x3  +  a;2  +  x  —  100  =  0. 

"When  4  and  5  are  substituted  for  x  in  the  equation,  the 
results  are  —  1G  and  +55.     Hence  we  have 

1G  +  55  :  5  -  4  : :  1G  :  x  -  4  ;  .-.     x  =  4.22. 

As  the  root  is  greater  than  4.2,  we  now  assume  4.2  and  4.3  ; 
substituting  these  for  x  in  the  given  equation,  aud  proceeding 
as  before,  we  obtain  for  a  second  approximation  x  =  4.264. 

Assuming  4.2G4  and  4.2G5  for  as,  and  repeating  the  oper- 
ation, we  obtain  for  a  third  approximation  u;  =  4.2644299 
nearly. 

2.  Find  a  root  of  a;3  +  30a  -  420  =  0.     Ana.  G.  170103. 


EXAMPLES. 


EXAMPLES. 


539 


Solve  the  following  equations : 

1.  x3  -  3a;2  -  46a;  -  72  =  0.  Ans.  9,  -2,  -4. 

2.  xa  -  2x2  —  4x  +  8  =  0.  2,  2,  -2. 

3.  X4  +  4x3  _  X2  _  16aJ  _  12  =  0.       2,  -1,  -2,  -3. 

4.  a;4  -  27a;2  +  14a;  +  120  =  0.  3,  4,  -2,  -5. 

5.  x4  -  12a;3  +  47a;2  -  72a;  +  36  =  0.  1,  2,  3,  6. 

6.  ^  _  9x3  +  l7x2  +  27a;  -  60  =  0.  4,  5,  ±^3. 

7.  a;5  +  5a;4  +  x3  -  16a;2  -  20a;  -  16  =  0.  

Ans.  2,  -2,  -4,  ~1  ^ 

Solve  the  following  reciprocal  equations  : 

8.  a;4+5a;3+2a;2+5a;+l  =  0.    Ans. $(-5±V^T),  ±V-1. 

9.  a;4  -  3a;3  +  3a;  -  1  =  0.  ±1,  £(3  ±  V^5). 

10.  a;4  -  10a;3  +  26a;2  -  10a;  +  1  =  0.     3 ±2^2,  2 ±^3. 

11.  4a;4  -  4a;3  -  7a;2  -  4a;  +  4  =  0.     2,  |,  i(-3±v/-7). 

12.  9a;4-24a;3-2a;2-24a;  +  9  =  0.     3,  A,  £(-l±\^-8. 

13.  a;6  -  11a;4  +  17a;3  +  17a;2  -  11a?  +  1  =  0. 

Ans.  -1,  |(9  ±  ^77),  |(3  ±  \£). 

14.  x5  -  5a;4  +  9a;3  -  9a;2  +  5a;  -  1  =  0. 

Ans.  1,£(1  ±  /^3),$(3±  ^5). 

15.  4a;6  -  24a;6  +  57a;4  -  73a;3  +  57a;2  -  24a;  +  4  =  0. 

Ans.  2,  2,  i,},  -1(1  ±  V/-3). 

16.  a;5  +  1  =  0. 

^s.  -1,  i[l+V/5±V/(+2V/5-10)],  i[l-V/5±V(_2V/5-10)]. 

17.  a;8  -  1  =  0.  ±!»  ±^1,  ±4  ±  N^« 

Solve  the  following  cubic  equations  : 

18.  a;3  +  63a;  -  316  =  0.  Ans.  4,  -2  ±  5V^3. 

19.  a;3  +  21a;  +  342  =  0.  -6,  3  ±  4V^3. 


540  EXAMPLES. 

20.  xs  -  6a;2  +  13a;  -  10  =  0.  Ans.  2,  2  ±  V^l. 

21.  a;3  +  6a;2  -  32  =  0.  2,  -4,  -4. 

22.  a;3  -  Gar  +  3a;  -  18  =  0.  6,  +V^3,  -V^3. 

23.  xs  -  15a;2  -  33a;  +  847  =  0.  11,  11,  7. 

24.  28x3  -  9x2  +1  =  0.  -£,  i(2  ±  V^). 

25.  2a;3  +  25a;2  +  56.x  -  147  =  0.  -7,-7,  1$. 

Find  a  root  by  Horner's  method  in  the  following  equations  : 

26.  x3  +  3z2  +  5a;  -  178  =  0.  Ans.  4.5388. 

27.  x3  +  11a;2  -  102a;  +  181  =  0.  3.21312. 

28.  a;3  +  x2  -   500  =  0.  7.61728. 

29.  a;3  +  a;2  +  x  -  100  =  0.  4.26443. 

30.  a;3  +  10a;2  +  6a;  -  120  =  0.  2.8330665. 

31.  a;4  -  8a;3  +  20a;2  -  15a;  +.5  =  0.  1.284724. 

32.  a;4  +  a;2  -  8a;  -  15  =  0.  2.302775. 

33.  a;5  +  4a;4  -  3a;8  +  10a;2  -  2a;  -  962  =  0.     3.385777. 

34.  a;4  -  2a;3  +  21a;  -  23  =  0.  1.1574515. 

35.  a;4  -  5a;3  +  3a;2  +  35a;  -  70  =  0.               2.6457513. 

Find  a  root  by  Newton's  method  in  the  following  equa- 
tions : 

36.  a;3  +  12a;2  -  18a;  -  216  =  0.  4.2426407. 

37.  x3  +  a;  -  3  =  0.  1.21341166. 

Find  a  root  by  the  method  of  Double  Position  in  the  fol- 
lowing equations : 

38.  a;3  +  2x  -  20  =  0.  2.46954565. 

39.  a;5  +  10a;2  +  8a;  -  120  =  0.  2.76834546. 


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l^evieW    Number    CardS.      By  Ella  M.  Pierce,  of  Providence,  R.  I 

For  Second  and  Third  Year  Pupils.     Cards,  7x9  inches 
per  hundred. 

Picture  Problems.    By  miss  h.  a.  luddington, 


For  Second  and  Third  Year  Pupils.     Cards,  7x9  inches.     Price,  3  cents  each ;  or  J3.40 
per  hundred. 


Principal  of  Training  School,  Pawtucket,  R.  I.  ;  formerly  Teacher  of  Methods  and  Train- 
ing Teacher  in  Primary  Department  of  State  Normal  School,  New  Britain,  Conn., 
and  Training  Teacher  in  Cook  County  Normal  School,  Normal  Park,  111.  70  colored 
cards,  4x5  inches,  printed  on  both  sides,  arranged  in  9  sets,  6  to  10  cards  in  each  set, 
with  card  of  directions.     Retail  price,  65  cents. 


{Mathematical  Teaching  and  its  [Modern  [Methods 

By  Truman   Henry   Saffc 
Mass.    Paper.     47  pages.    I 

The  New  Arithmetic. 


By  Truman   Henry   Safford,  Ph.  D.,   Professor  of   Astronomy,   Williams   College, 
Mass.     Paper.     47  pages.    Retail  price,  25  cents. 


By  300  authors.     Edited  by  Seymour  Eaton,  with  Preface  by  T.  H.  Safford,  Pr» 
fcssor  of  Astronomy,  Williams  College,  Mass.     Introduction  price,  75  cents. 


D.    C.    HEATH    &    CO.,    Publishers, 

BOSTON,   NEW  YORK,   AND  CHICAGO. 


Hnglish   Language. 


Hyde's  Lessons  in  English,  Book  I.  For  the  lower  grades.  Con- 
tains exercises  for  reproduction,  picture  lessons,  letter  writing, 
uses  of  parts  of  speech,  etc 5-35 

Hyde's  Lessons  in  English,  Book  II.       For  Grammar  schools. 

Has  enough  technical  grammar  for  correct  use  of  language  .        .50 

Hyde's  Lessons  in  English,  Book  II.  with  Supplement.     Has 

in  addition  to  the  above,  118  pages  of    technical  grammar        .        .60 
Supplement  bound  alone 30 

Hyde's  Derivation  of  Words  .  .....       .15 

Mathew's  English  Grammar  with  Selections 70 

3uckbee's  Primary  Word  Book 25 

Badlam's  Suggestive  Lessons  in  Language.     Being  Part  I.  and 

appendix  of  Suggestive  Lessons  in  Language  and  Reading   .         .        .50 

Smith's  Studies  in  Nature,  and  Language  Lessons.  A  combi- 
nation of  object  lessons  with  language  work  .50  Part  I  bound 
separately 25 

Meiklejohn's  English  Language.     Treats  salient  features  with  a 

master's  skill  and  with  the  utmost  clearness  and  simplicity  .         .      1.20 

Meiklejohn's  English  Grammar.   Also  composition,  versification, 

paraphrasing,  etc.      For  high  schools  and  colleges         ...        .80 

Meiklejohn's  History  of  the  English  Language.     78  pages.    Part 

III.  of  English  Language,  above  . 30 

Williams'   Composition   and  Rhetoric   by   Practice.     For  high 

school  and  college.  Combines  the  smallest  amount  of  theory  with 

an  abundance  of  practice.     Revised  edition 90 

Strang's  Exercises  in  English.     Examples  in  Syntax,  Accidence, 

and  Style  for  criticism  and  correction.     Revised  edition      .         .         .45 

Hempl's  Old  English  Grammar  and  Reader 25 

Huffcutt's  English  in  the  Preparatory  School.  Presents  as  prac- 
tically as  possible  some  of  the  advanced  methods  of  teaching  Eng- 
lish grammar  and  composition  in  the  secondary  schools         .         .        .25 

Woodward's  Study  of  English.     Discusses  English  teaching  from 

primary  school  to  high  collegiate  work 25 

Genung's  Study  of  Rhetoric.    Shows  the  most  practical  discipline 

of  students  for  the  making  of  literature 25 

In  addition  to  the  above  we  have  text-books  in  English  and  American 
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D.  C.  HEATH  &  CO.,  PUBLISHERS. 

BOSTON,  NEW   YORK  &  CHICAGO. 


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